Answer:
the height at which the rock's kinetic energy is twice its potential energy is approximately 4.5 m.
Explanation:
When the rock is dropped from a height of 9.0 m, it initially has potential energy equal to mgh, where m is the mass of the rock, g is the acceleration due to gravity (9.8 m/s^2), and h is the height from which the rock was dropped. Therefore, the potential energy of the rock is:
U = mgh = (6.0 kg) * (9.8 m/s^2) * (9.0 m) = 529.2 J
As the rock falls, its potential energy is converted to kinetic energy, given by the expression (1/2)mv^2, where v is the velocity of the rock. At a height where the kinetic energy of the rock is twice its potential energy, we can write:
(1/2)mv^2 = 2mgh
Simplifying this expression, we get
v^2 = 4gh
At this height, the kinetic energy of the rock is given by:
K = (1/2)mv^2 = (1/2)m(4gh) = 2mgh = 2U
Substituting the values of m, g, and U, we get:
v^2 = 4gh = 4(9.8 m/s^2)h = (2 * 529.2 J) / 6.0 kg = 176 J/kg
Solving for h, we get:
h = (v^2) / (4g) = (176 J/kg) / (4 * 9.8 m/s^2) ≈ 4.5 m
A hockey puck slides along a horizontal, smooth icy surface at a constant velocity as shown.(Figure 1) Which of the following forces act on the puck?
weight
force of push
air drag
force of velocity
acceleration
friction
normal force
The weight and normal force are the only two forces acting on the hockey puck vertically; however, because the puck is travelling at a constant speed, there are no forces acting on it horizontally.
What does normal force mean in everyday language?A force known as the "normal force" is felt when an object is placed on a surface and pressed against it by that surface.
Provide a definition of normal force and a scenario where a book is put on a table:We come into contact with the usual force every day. When we place a book on a table, for instance, the usual reaction force stops it from falling through.
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The two boxes are sliding along a frictionless surface. 4 kg box is moving right at 9m/s and 1.5kg box is moving left at 24m/s. They collide and stick
together. Afterward, the velocity of the two boxes is
The velocity of the two boxes after the collision is 0 m/s, which means they come to a complete stop.
option B.
What is the final velocity of the two boxes?We can solve this problem by applying the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.
Before the collision, the momentum of the 4 kg box is:
p1 = m1 * v1 = 4 kg * 9 m/s = 36 kgm/s (to the right)
Before the collision, the momentum of the 1.5 kg box is:
p2 = m2 * v2 = 1.5 kg * (-24 m/s) = -36 kgm/s (to the left)
The total momentum before the collision is:
p_total = p1 + p2 = 36 kgm/s - 36 kgm/s = 0 kgm/s
The total momentum after the collision is also 0 kgm/s, since there are no external forces acting on the system.
p_total = m_total * v_final
where;
m_total is the total mass of the two boxes, and v_final is their common velocity after the collision.We can solve for v_final:
v_final = p_total / m_total = 0 kg*m/s / (4 kg + 1.5 kg) = 0 m/s
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in fig. 8-29, a 2.30 g ice flake is released from the edge of a hemispherical bowl whose radius r is 35.0 cm. the flake-bowl contact is frictionless.
The flake will travel downwards in a parabolic path in a curved trajectory before hitting the bottom of the bowl, due to the force of gravity acting on it. No friction is present so the final velocity of the flake will be the same as its initial velocity.
What is Velocity?Velocity is a measure of the rate and direction of change in the position of an object. It is a vector quantity that is equal to the rate of change of the object’s position in a given direction. Velocity is expressed as a vector, which is composed of both a magnitude and a direction.
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The earth's crust is made of tectonic plates. These plates are constantly in motion. How are earthquakes related to the movement of tectonic plates?.
Earthquakes are related to the movement of tectonic plates, as the plates' motion can cause stress to build up in the crust, leading to sudden movements along faults, resulting in earthquakes.
Tectonic plates are constantly moving, driven by convection currents in the mantle. As plates move, they can rub against each other, causing friction and stress to build up along their boundaries. This stress can eventually overcome the strength of the rocks, resulting in sudden movements along faults, which are zones of weakness in the crust.
These sudden movements release energy in the form of seismic waves, causing the ground to shake, resulting in an earthquake. Therefore, earthquakes are directly related to the movement of tectonic plates and occur most frequently along plate boundaries.
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The Origin of Magma Draw arrows on simplified Pressure/Temperature (P/T) diagrams according to the scenarios below. Be sure to start your arrow at the red dot and draw it in the direction the dot would move in each scenario. 1. An area of solid rock in the subsurface is heated quickly in place and starts to melt. Indicate the direction the dot would move in P/T space. 2. An area of solid rock undergoes increases in pressure and temperature as it is buried in the subsurface, following parallel to the liquid/solid line for this exercise. Indicate the direction the dot would move in P/T space. Temperature (°C) Temperature (°C) 1. Low High Low 2. High Pinbr 3. An area of solid rock deep in the subsurface quickly rises to a lower depth/pressure and melts. Indicate the direction the dot would move in P/T space. At which type of plate boundary does this type of melting occur? 4. Magma erupts at the surface and then rapidly cools to a solid. Indicate the direction the dot would move in P/T space. Temperature (°C) Temperature (°C) Low 3. High Low 4. High 5. If water is added to the system, which way does the liquid/solid line move? Draw in the new solidus line. At which type of plate boundary does this type of melting occur? 6. Dotted line is the liquid/solid line of a wet system. If water is added to a portion of dry rock (dot), will it melt in place? Label the dot 'Melt' or 'No Melt' to indicate your answer. Temperature (C) Temperature (°C) 5. Low High Low High 6. Pressure (bars x 1,000) Depth (km) Pressure (bars x 1,000) Depth (km) Examine the pressure-temperature (P-T) diagram below and locate point X. This point represents a piece of peridotite (the type of rock that exists in the mantle) buried 80 km underground. Ocean 0 0 500 TEMPERATURE IN DEGREES CELSIUS (°C) 1000 1500 2000 2500 Crust 10,000 Lithosphere 50 100% LIQUID MAGMA 20,000 Lithospheric Mantle 100- 30,000 Decompression 40,000 DEPTH IN KILOMETERS (km) Continental geothermal gradient Oceanic geothermal gradient 150- Peridotite solidus PRESSURE IN ATMOSPHERES (atm) 50,000 60.000 Asthenospheric Mantle 200- 100% SOLID PERIDOTITE ROCK PARTIAL MELTING OF PERIDOTITE ! (Solid + Liquid) 1 70,000 250 80,000 Heating Melting 1 1 90,000 300 A. Earth's Mechanical Layers B. Pressure-Temperature Diagram 0 25% 50% 75% 100% 1. According to the continental geothermal gradient, rocks buried 80 km beneath a continent would normally be heated to what temperature? 2. According to the oceanic geothermal gradient, rocks buried 80 km beneath an ocean basin would normally be heated to what temperature? 3. Is the mantle rock at point X a mass of solid, a mixture of solid and liquid, or a mass of liquid? How do you know? 4. What would happen to the mass of mantle at point X if it were heated to 1750 °C? 5. What would happen to the mass of mantle at point X if it were heated to 2250 °C? 6. At its current depth, the mantle at point X is under about 25,000 atm of pressure. a. At what depth and pressure will this rock begin to melt if it is uplifted closer to Earth's surface and its temperature remains the same? Depth: Pressure: b. What is the name applied to this kind of melting? At which plate boundary setting does the uplifting of mantle rock cause melting in this way? 7. Based on your answers above, what are two environmental changes that can cause the mantle at point X to begin partial melting?
Two environmental changes that can cause the mantle at point X to begin partial melting are heat and pressure reduction.
What is the pressure reduction?Pressure reduction is a process that occurs when the pressure of an object or material is decreased. This can be done in a variety of ways, such as by using a pressure-reducing valve, increasing the area of a pipe, or using a compressor to reduce the pressure of a gas or liquid.
1. Low High Low
2. High Pinbr
3. Low High Low
4. High Low
5. The liquid/solid line would move to the right, indicating an increase in solidus temperature. This type of melting occurs at subduction zones.
6. No Melt
1. According to the continental geothermal gradient, rocks buried 80 km beneath a continent would normally be heated to about 1400 °C.
2. According to the oceanic geothermal gradient, rocks buried 80 km beneath an ocean basin would normally be heated to about 1000 °C.
3. The mantle rock at point X is a mass of solid, as indicated by the fact that it is below the solidus line.
4. If heated to 1750 °C, the mass of mantle at point X would remain a solid.
5. If heated to 2250 °C, the mass of mantle at point X would begin to partially melt.
6. a. At its current temperature, the mantle at point X will begin to melt at a depth of about 50 km and a pressure of about 10,000 atm.
b. The name applied to this kind of melting is decompression
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An astronomy textbook, when printed out, weighs four pounds on the surface of the Earth. After finishing your course, you are so tired of the book, you arrange for NASA to shoot it into space. When it is twice as far from the center of the Earth than when you were reading it, what would it weigh? (Note, assume that the book has been moving away from the Earth, not falling freely around it.)
The book will weigh 1 pound when it is twice as far from the center of the Earth as when it was on the surface.
Weight of objects in spaceThe weight of an object depends on its mass and the gravitational force acting on it. Near the surface of the Earth, the gravitational acceleration is approximately 9.8 m/s^2. We can use the inverse square law to calculate the gravitational force on the book when it is twice as far from the center of the Earth than when it was on the surface:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant, m1 is the mass of the Earth, m2 is the mass of the book, and r is the distance between the center of the Earth and the book.
Since the mass of the book remains the same, the weight of the book on the surface of the Earth is simply the force of gravity acting on it:
W = m * g
where W is the weight of the book, m is its mass, and g is the gravitational acceleration near the surface of the Earth.
When the book is twice as far from the center of the Earth, the gravitational force on it is:
F' = G * (m1 * m2) / (2r)^2 = F / 4
Therefore, the weight of the book when it is twice as far from the center of the Earth is:
W' = (m * g) * (F' / F) = (m * g) * (1/4) = 1 pound
So the book would weigh one pound when it is twice as far from the center of the Earth than when it was on the surface.
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HELPP
im not exactly sure what to put
Explanation:
between 0 and 2 hours the relationship between time and distance traveled is a linear one.
that means that the velocity (= distance/time like in km/s) is constant (and not increasing).
and because between the hours 2 and 3 there is no increase in distance, this means the car stood still during that time. so, no, it was not always in motion during the voyage.
therefore,
option 2 is correct (as described, no change in distance means no motion or zero velocity).
and option 3 is correct, because also between hours 3 and 5 the function is linear and the speed ratio is therefore constant.
What is determined by calculating the slope of the position vs time graph? position velocity distance displacemen
A piece of iron has a weight of 3. 5 n when it is in air and 2. 0 n when it is submerged into water. What is the buoyant force on the piece of iron?.
The buoyant force acting on the piece of iron is 1.5 N.
The weight of an object in air is equal to the force of gravity acting on the object. The weight of an object in water is equal to the force of gravity minus the buoyant force acting on the object.
Therefore, we can calculate the buoyant force on the piece of iron as follows:
Weight of iron in air = 3.5 N
Weight of iron in water = 2.0 N
Buoyant force = Weight of iron in air - Weight of iron in water
Buoyant force = 3.5 N - 2.0 N
Buoyant force = 1.5 N
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A flat puck (mass M) is rotated in a circle on a frictionless air-hockey tabletop, and is held in this orbit by a light cord connected to a dangling block (mass m) through a central hole in the table as shown in the figure. Start by showing the free-body diagrams and force equations for each mass, and then derive an equation for the tangential speed of the puck. This will be an algebraic expression that includes the variables mentioned above.
The dangling mass will not move and the puck will keep rotating in its orbit, Therefore, V = √(mgR/M) will be an algebraic expression that includes the variables mentioned above.
What is centripetal force?A centripetal force that causes a body to travel along a curved path. The centripetal force always acts orthogonally to the motion of the body and towards the fixed point of the path's instantaneous centre of curvature. Isaac Newton described it as "a force by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre". Gravity is the centripetal force that causes astronomical orbits, according to Newtonian mechanics theory.
A common example of centripetal force is when a body moves at a constant speed along a circular path. The centripetal force is directed at right angles to the motion as well as along the radius of the circular path towards the centre.
Fc = mg
Now, centripetal force on the puck is equal to the mass of the puck times the centripetal acceleration, or
Fc = Mac
centripetal acceleration is related to the tangential velocity of the puck as
ac = V²/R
where R is the radius of the orbit. Then we have
Fc = M ac =M ω² R = M V² / R
and, thus
M V² / R = mg
and, solving for V, we get
V = √(mgR/M)
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An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 24 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination?.
The plane must fly at an angle of 11.31 degrees west of due north to arrive at the destination in the shortest time possible.
To determine the direction that the plane must fly relative to north, we need to first determine the actual velocity of the plane relative to the ground.
Let's break down the velocity vectors involved in the flight:
The air speed of the plane is 120 m/s in a direction perpendicular to the plane's heading (i.e., to the east).
The wind speed is 24 m/s due west.
Using vector addition, we can find the resultant velocity vector of the plane relative to the ground:
The eastward component of the plane's velocity is 120 m/s.
The westward component of the wind's velocity is 24 m/s.
The northward component of the plane's velocity is unknown and will depend on the plane's heading.
We can use the Pythagorean theorem to find the magnitude of the resultant velocity vector:
[tex]resultant speed^2 = eastward speed^2 + northward speed^2[/tex]
[tex]resultant speed^2 = (120 m/s)^2 + northward speed^2[/tex]
resultant speed = [tex]\sqrt{[(120 m/s)^2 + northward speed^2]}[/tex]
Since the destination is directly north of the origin, the plane's heading must be northward. Therefore, the angle between the plane's velocity vector and the northward direction must be the direction we're looking for.
To find this angle, we can use trigonometry. Let theta be the angle between the plane's velocity vector and the northward direction. Then:
tan(theta) = northward speed / 120 m/s
northward speed = 120 m/s * tan(theta)
Substituting this into the equation for the magnitude of the resultant velocity vector, we get:
resultant speed = [tex]\sqrt{[(120 m/s)^2 + (120 m/s * tan(theta))^2]}[/tex]
We want the resultant velocity vector to be equal to the distance between the two points on the ground (500 km) divided by the time of the flight. We can convert this to meters per second by dividing by the duration of the flight in seconds:
resultant speed = 500000 m / (flight time in seconds)
Equating this with the expression we derived for the magnitude of the resultant velocity vector, we get:
500000 m / (flight time in seconds) = [tex]\sqrt{[(120 m/s)^2 + (120 m/s * tan(theta))^2]}[/tex]
Solving for the flight time in seconds and simplifying, we get:
flight time = [tex]500000 m \sqrt{[(120 m/s)^2 + (120 m/s * tan(theta))^2]}[/tex]
To minimize the flight time, we need to maximize the northward component of the plane's velocity. We can do this by making the angle theta as small as possible, i.e., by flying as close to due north as possible.
Taking the derivative of the flight time expression with respect to theta, we get:
d(flight time) / d(theta) =[tex]-60000 * tan(theta) / (120^2 * (1 + tan(theta)^2)^(3/2))[/tex]
Setting this equal to zero, we find that the optimal value of theta is arctan(1/5), which is approximately 11.31 degrees. Therefore, the plane must fly at an angle of 11.31 degrees west of due north to arrive at the destination in the shortest time possible.
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1. Bone has a Young’s modulus of about 1.8 × 10^10 Pa. Under compression, it can
withstand a stress of about 1.61 × 10^8 Pa before breaking. Assume that a femur (thigh bone) is 0.47 m long, and calculate the amount of compression
this bone can withstand before breaking.
Answer in units of mm.
2 (Pt 1/2). How much pressure is applied to the ground by a 68 kg man who is standing on square stilts that measure 0.04 m on each edge?
Answer in units of Pa.
3 (Pt 2/2). What is this pressure in pounds per square inch?
Answer in units of lb/in^2.
4 (Pt 1/2). In an 84 s interval, 578 hailstones strike a glass window of area 0.828 m^2 at an angle 32° to the window surface. Each hailstone has a mass of 5 g and speed of 11.7 m/s. If the collisions are elastic, find the average force on the window.
Answer in units of N.
5 (Pt 2/2). Find the pressure on the window.
Answer in units of N/m^2.
6. If a 1-megaton nuclear weapon is exploded at ground level, the peak overpressure (that is, the pressure increase above normal atmospheric pressure) will be 0.2 atm at a distance of 6 km. Atmospheric pressure is 1.013 × 10^5 Pa.
What force due to such an explosion will be exerted on the side of a house with dimensions 2.09 m × 30.9 m?
Answer in units of N.
7. Find the density of seawater at a depth where the pressure is 130 atm if the density at the surface is 1050 kg/m^3. Seawater has a bulk modulus of 2.3 × 109 N/m^2. Bulk modulus is defined to be (look at picture).
Answer in units of kg/m^3.
8. Calculate the depth in the ocean at which the pressure is three times atmospheric pressure. Atmospheric pressure is 1.013 × 10^5 Pa. The acceleration of gravity is 9.81 m/s^2 and the density of sea water is 1025 kg/m^3.
Answer in units of m.
Answer:
1. Y = [tex]\frac{P*L}{Y}[/tex] = ((1.61 × 10⁸) × 0.47) ÷ 1.8 × 10¹⁰
Y = 4.20Ε17 mm
2. P = F/A
F = m×g = 68 ×9.8 = 666.4N
A = (2 · (0.04)² = 0.0032²
P = 666.4 ÷ 0.0032 = 208250 Pa
3. Psi = 208250 ÷ 6894.76
Psi = 30.20 lb/in²
4. Δp = 2 × mass × velocity
Δp = 2 × 0.005 × 11.7 × sin32 = 0.0620 kgm/s
Δp = 578 × 0.0620 = 35.83 kgm/s
F = Δp/Δt = 35.83 ÷ 84
F = 0.4266N
5. P = F/A = 0.4266 ÷ 0.828
P = 0.515244 N/m²
**(I'll add the last few later)**
if the net force of F is applied to half the mass (m/2), write the acceleration of the mass in terms of a.
The acceleration of the mass in terms of {m} is -
a = 2F/m.
What is the formula to calculate the force acting on a body?The force acting on a body is given by -
Force {F} = mass {m} x acceleration {a}
Given is that a net force of {F} is applied to half the mass {m/2}.
We know that -
Force {F} = mass {m} x acceleration {a}
F = {m/2} x a
a = 2F/m
Therefore, the acceleration of the mass in terms of {m} is -
a = 2F/m.
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Now Jolon and Terry tackle a problem: A block of mass 2.0 kg is attached to a horizontal spring that has a force constant of 2.90 X 103 N/m, and is free to slide on a frictionless surface as shown: The spring is compressed to Xj -6.5 cm by pushing on the block, and then the block is released_ Find the work done by initially compressing the spring: Find the kinetic energy of the block when it reaches x = 0_ Find the speed of the block at x = 0. m/s
The speed of the block at x = 0 is 1.84 m/s.
What is kinetic energy?Kinetic energy is a type of energy that an object possesses due to its motion. It is defined as the energy that an object has by virtue of its motion, and is given by the equation:
K = (1/2)mv²
where K is the kinetic energy of the object, m is its mass, and v is its velocity.
We use the conservation of energy principle. Initially, the block has potential energy stored in the spring, which is converted into kinetic energy as the block moves towards its equilibrium position, where x=0.
The potential energy stored in the spring when it is compressed to x = -6.5 cm is given by:
U = (1/2)kx²
where k is the force constant of the spring and x is the displacement from its equilibrium position.
Substituting the given values, we get:
U = (1/2) (2.90 X 10³ N/m) (0.065 m)²
U = 6.73 J
Therefore, the work done by compressing the spring is 6.73 J.
At the equilibrium position, all the potential energy stored in the spring is converted into kinetic energy. So, kinetic energy of the block at x = 0 is equal to potential energy stored in spring at x = -6.5 cm.
Therefore, the kinetic energy of the block at x = 0 is also 6.73 J.
Speed of the block at x = 0:
The kinetic energy of the block at x = 0 is given by:
K = (1/2)mv²
where m is the mass of the block and v is its speed.
Substituting the given values, we get:
6.73 J = (1/2) (2.0 kg) v²
Solving for v, we get:
v = √((2 x 6.73 J) / 2.0 kg)
v = 1.84 m/s
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On a particular day, the following appliances are used for the times indicated: a 1600-W coffee maker, 13 min, and a 1100-W microwave oven, 4.5 min. With these power requirements, find how much it costs to use these appliances at an electrical cost of 7¢ per kWh. (Do not round your final answer. Fractional cent values are acceptable.)
The cost of using the appliances will be 3004.19¢
What is electricity cost?The electricity cost is defined as the amount of money required to use per Kw of electric power.
Given that on a particular day, the following appliances are used for the times indicated: a 1600-W coffee maker, 13 min, and an 1100-W microwave oven, 4.5 min.
The first step is to convert the times from minutes to hours.
60 min = 1 hour
13 min = 13 /60
4.5 min = 4.5/60
Considering the 1600-W coffee maker,
The total consumption will be calculated as:-
Consumption = 1600 x 13 /60 = 346.67 kWh
Considering the 1100-W microwave oven,
The total consumption will be calculated as:-
Consumption = 1100 x 4.5/60 = 82.5 kWh
Total consumption = 346.67 + 82.5 = 429.17 kwh
Given that the cost is 7¢ per kWh,
Cost of using these appliances = 429.17x 7
Cost of using these appliances = 3004.19 ¢
Therefore, the cost of using the appliances will be 3004.19¢
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solve it and the answer is in J
The final answer, in joules, is -1835.120716 J.
What is radian?Angles are measured in terms of radians. It is described as the angle occupied by a circle arc whose length is equal to the circle's radius. In other words, a circle with a radius of 1 unit subtends an arc with a length of 1 unit at an angle of 1 radian. Radians have no units because they are a dimensionless quantity.
Let's simplify and evaluate the given expression step by step:
First, let's calculate the terms inside the brackets:
(83)(2.1)²/4 = 185.3675 J
(100)(0)²/2 = 0 J
So, the first term in the expression becomes:
185.3675 J
Now, let's calculate the second term inside the brackets:
(83)(2.1)²/4 = 185.3675 J
(100)(2.1)²/2 = 2205 J
(0.14 rev/s * 2π rad/rev)² = 0.246784 J/(rad^2)
So, the second term in the expression becomes:
185.3675 J + 2205 J - 0.246784 J = 2390.120716 J
Now, we can calculate the final expression by plugging in the values and simplifying:
[(83)(2.1)²/4 +(100)(0)²/2] * (3 rad/sec)² - [(83)(2.1)²/4 +(100)(2.1)²/2 * (0.14 rev/s)² (2pi rad/rev)²]
= [185.3675 J + 0 J] * (3 rad/sec)² - [185.3675 J + 2205 J - 0.246784 J]
= 555 J - 2390.120716 J
= -1835.120716 J
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QUESTION 4 A motorcycle is traveling at a speed of 10 km/h along a straight road. Express the speed in m/s? (2 MARKS]
Answer:
≈ 3 m/s
Explanation:
To express the speed which is given in km/h to m/s, we can simply individually change from km to m and h to s.
We know that 1 kilometer = 1000 meters and that 1 hour = 3600 seconds. Therefore:
[tex]\displaystyle{10 \cdot \dfrac{1000 \ \: \text{m} }{3600 \ \: \text{s}}}\\\\\displaystyle{=10 \cdot \dfrac{10 \ \: \text{m} }{36\ \: \text{s}}}\\\\\displaystyle{=\dfrac{100 \ \: \text{m} }{36 \ \: \text{s}}}\\\\\displaystyle{\approx 3 \ \: \text{m/s}}[/tex]
Therefore, 10 km/h will approximately equal 3 m/s as accorded to significant rules.
Derive Lorentz transformation
Lorentz transformation is a mathematical framework that describes how the coordinates of an event in space and time change for two observers in relative motion.
What is Lorentz transformation?Consider two inertial frames of reference, S and S', moving relative to each other along the x-axis with a constant velocity v. Let an event occur at position (x, y, z, t) in frame S, and let (x', y', z', t') be the corresponding position in frame S'.
The transformation equations for space and time are given by:
[tex]x' = γ(x - vt)\\y' = y\\z' = z\\t' = γ(t - vx/c^2)[/tex]
where[tex]γ = 1/√(1 - v^2/c^2)[/tex] is the Lorentz factor, and c is the speed of light.
These equations show how the coordinates of an event in one frame of reference are related to the coordinates in another frame of reference that is moving relative to the first frame. They are known as the Lorentz transformations and are a fundamental aspect of special relativity.
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A man pushes a lawn mower on a level lawn with a force of 207 N. If 41% of this force is directed forward, how much power does he expend in pushing the mower 4.8 m in 18 s?
The power which is required to push the mower on a level lawn with a force of 207 N will be 32.568 watts.
What is Power?Power can be defined as the rate of work done per unit of time taken to move or displace an object from one location to another location.
When considering work done on the object, we always take the force directed along the axis of motion, which is in this case, the horizontal axis. If 59% of the force is directed downward, then 41% of the force is being directed horizontally, so the horizontal force is
207 × 0.59 = 122.13N,
Work done = Force applied × displacement of the object
Work done = 122.13 × 4.8 = 586.22 Joules
Power = Work done/ Time taken
Power = 586.22/ 18
Power = 32.568 watts (W)
Therefore, the power will be 32.568 watts.
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A 6.0-kg rock is dropped from a height of 9.0 m. At what height is the rock's kinetic energy twice its potential energy?
A rock weighing 1 kilogramme is thrown from a distance of six metres. height is the difference between the rock's kinetic and potential energy?
Describe energy?People have figured out how to transform energy from one type to another and use it to accomplish tasks, making modern civilization possible.
A possibility is what?In a wide range of disciplines, including physics and the social sciences, the phrase is used to denote objects that are capable of changing
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Resolving power(R.P) is a function of the wavelength of light used and the numerical aperture of the objective lens. It is expressed as the smallest distance between two points that can still be discerned as discrete entities.
Given that R.P. = wavelength/ (2* N.A) Calculate the resolving power of your microscope when using the oil immersion lens. Assume that visible light has an average wavelength of 0.55 um (a um= 0.000001 meters).
The resolving power of the microscope is approximately 0.196 μm when using the oil immersion lens.
What is Aperture?
In optics, the aperture refers to the opening or hole in an optical system through which light passes. It is typically a circular or rectangular opening that can be adjusted to control the amount of light that enters the system. The aperture size affects various properties of the optical system, such as the depth of field, the amount of light that reaches the sensor or film, and the resolving power or sharpness of the resulting image. The numerical aperture (N.A.) is a related term that describes the ability of an optical system to gather light, and it is determined by the size of the aperture and the refractive index of the medium through which light passes.
To calculate the resolving power (R.P.) of the microscope, we can use the formula:
R.P. = wavelength / (2 * N.A.)
where wavelength is the average wavelength of visible light, and N.A. is the numerical aperture of the objective lens.
Given that the average wavelength of visible light is 0.55 μm (micrometers), and assuming we are using an oil immersion lens, which typically has a numerical aperture of around 1.4, we can plug in these values and solve for R.P.:
R.P. = (0.55 μm) / (2 * 1.4)
R.P. = 0.196 μm
Therefore, the resolving power of the microscope is approximately 0.196 μm when using the oil immersion lens.
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A soccer ball is kicked at an angle of 45° above the horizontal and travels a horizontal distance of 15 meters. If soccer ball was kicked at the same speed but change the angle to 60°, then how far will it travel horizontally?
Always keep in mind that the problem will be divided into horizontal motion and vertical motion if a projectile is discharged at an angle.
Describe projectile?A projectile is any object that is sent into orbit with only gravity acting upon that. The projectile is mostly affected by gravity. This doesn't mean that other forces don't have an impact; it just means that they have a much smaller one compared to gravity.
We can determine the duration of flight by solving the vertical motion equations. The range is then determined by applying the equations for horizontal motion. We must resolve the system of both equations in your inquiry.
I utilise the following three kinematic motion equations to address practically any projectile motion issue:
According to the given data:S=Vit+12at2
————--eqn 1
Vf=Vi+at
—————eqn 2
combine the equation 1 and 2 to eliminate “t” gives
V2f−V2i=2aS
—————eqn 3
Always watch your signs in these equations. Velocities are up = positive, down = negative and the acceleration due to gravity always points down so ay=−9.81ms2.
Moreover, S is negative if indeed the finishing position is lower than the initial position. S=0 in response to your query because it lands at the identical height.
Vertical Motion:
Let's use an equation without final velocity as we typically don't know it:
S=Vit+12at2
I’ll add subscripts to indicate vertical direction:
Sy=(Vi)yt+12ayt2
0=(Vsin40)t+12(−9.81)t2
cancel “t” in all terms gives
0=Vsin40+12(−9.81)t
or
2Vsin40=9.81t
—— equation 1
Horizontal Motion:
Sx=(Vi)xt+12axt2
but air resistance is negligible so ax=0
Sx=(Vi)xt
25=(Vcos40)t
or
t=25Vcos40
—— equation 2
Substitute equation 2 into equation 1:
2Vsin40=9.81(25Vcos40)
or
V2=(9.81)(25)2sin40cos40
V=15.8ms
Maximum Height:
At maximum height, velocity = 0
(Vf)2y−(Vi)2y=2aySy
0−(15.78sin40)2=2(−9.81)Sy
Sy=5.24m
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reposting this question
a plane is flying north at 30 m/s. it is encountering wind blowing east at 5m/s. what is the airplane's speed?
Answer:
30.41 m/s
Explanation:
To solve this problem, we can use vector addition. We can represent the velocity of the plane as a vector pointing north with a magnitude of 30 m/s, and the velocity of the wind as a vector pointing east with a magnitude of 5 m/s.
The speed of the airplane is the magnitude of the resultant vector, which is the vector sum of the velocity of the plane and the velocity of the wind. To find the magnitude of the resultant vector, we can use the Pythagorean theorem:
magnitude of resultant vector = sqrt((30 m/s)^2 + (5 m/s)^2) = sqrt(900 m^2/s^2 + 25 m^2/s^2) = sqrt(925 m^2/s^2) = 30.41 m/s
Therefore, the speed of the airplane is approximately 30.41 m/s.
Which sequence correctly describes the route sperm take through the human male reproductive system?.
"The sequence that correctly describes the route male gamete take through the human male reproductive system is epididymis, vas deferens, urethra."Correct option is B.
The act of reproduction is how organisms create more of their own kind. There are two types of gametes used in human reproduction. The zygote undergoes a procedure in which it transforms into an embryo and then into a fetus.
The male has genitalia, which are reproductive structures, both inside and outside the pelvic girdle. The epididymis and vas deferens, which carry male gamete, are located next to the testicles. The scrotum, an external pouch-like structure, house the epididymis.
The seminal vesicles and the prostate gland are examples of accessory glands that produce secretions that lubricate the duct system and feed the male gamete. The male gamete travels from the genetalia of man through the urethra to the outside of the body.
Thus, the sequence that correctly describes the route male gamete take through the human male reproductive system is epididymis, vas deferens, urethra. Best choice is B.
The given question is incomplete. The complete question contains options. They are 'a. vas deferens, urethra, epididymis b. epididymis, vas deferens, urethra c. vas deferens, epididymis, urethra d. urethra, epididymis, vas deferens'
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A 65. 0 kg ice skater standing on frictionless ice throws a 0. 15 kg snowball horizontally at a speed of 32. 0 m/s. At what velocity does the skater move backward?.
The skater moves backward with a velocity of approximately 0.074 m/s.
We can calculate using momentum formula.
According to the law of conservation of momentum, the total momentum of a system is conserved if there is no external force acting on the system. In this case, the system consists of the ice skater and the snowball, and the external force acting on the system is assumed to be negligible. Therefore, the initial momentum of the system must be equal to the final momentum of the system.
The initial momentum of the system is given by the product of the mass of the skater and the initial velocity of the skater, which is zero, since the skater is initially at rest. The momentum of the snowball is given by the product of its mass and its initial velocity.
[tex]pi = ms * vs + mb * vb[/tex]
where pi : initial momentum of the system, ms : mass of the skater, vs is : initial velocity of the skater, mb : mass of the snowball, and vb : initial velocity of the snowball.
The final momentum of the system is given by the product of the mass of the skater and the final velocity of the skater, which is what we need to find, and the momentum of the snowball, which is equal to its mass times its final velocity, since it stops moving after it is thrown.
[tex]pf = ms * vf + mb * 0[/tex]
pf : final momentum of the system, vf : skater's final velocity, and 0 : final velocity of the snowball.
Since the initial momentum of the system is equal to the final momentum of the system, we can write:
[tex]ms * vs + mb * vb = ms * vf[/tex]
Substituting:
65.0 kg * 0 m/s + 0.15 kg * 32.0 m/s = 65.0 kg * vf
vf = -0.074 m/s
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What speed will the charge have after accelerating through a potential difference equal to 4v?.
The speed of the charged particle q after accelerating through a potential difference of 4V is given by sqrt(8qV/m).
The kinetic energy gained by a charged particle q accelerated through a potential difference V is given by:
K = qV
If the potential difference is increased to 4V, the kinetic energy gained by the particle will be:
K' = q(4V) = 4qV
Since the particle starts from rest, all the energy gained is converted to kinetic energy. Therefore, equating K' to the kinetic energy of the particle, we get:
K' = 1/2 mv^2
where m is the mass of the particle and v is the final speed of the particle.
Equating the two equations, we get:
4qV = 1/2 mv^2
Solving for v, we get:
v = sqrt(8qV/m)
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The given question is incomplete, the complete question is:
Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after accelerating through potential difference 4V?
In both cases shown, a box is sliding across a floor with the same coefficient of kinetic friction and the same initial velocity The only difference between the two cases is the mass of the box. In which case will the box slide the furthest before coming to rest? a. Case 1 b. Same c. Case 2
The more frictional force is acting on greater mass. Hence, less it will be accelerated to significant distance. So that in case 1 with small mass will slide farthest.
What is friction ?Friction is the resistive force that hinder the motion of an object. It just opposes the normal force of an object. Thus, frictional force will have a negative sign always.
The force of exerted on an object is directly proportional to its mass. Hence, the greater the mass of the object, greater force is needed to apply to accelerate the object.
Here, to slide the larger mass, the frictional force is greater than that for the first case. However, the small mass can be moved to slide the surface furthest. Hence case 1 is correct.
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A 117-lb student races up stairs with a vertical height of 5.7 m in 5.5 s to get to a class on the second floor. How much power in watts does the student expend in doing work against gravity?
A 117-lb student races up stairs with a vertical height of 5.7 m in 5.5 s to get to a class on the second floor. The power required to doing work against the gravity is 538.28 watts.
What is Power?Power is the amount of energy transferred per unit time. The SI unit of power is the watt, which is equal to one joule per second. Power is sometimes also called as activity. Power is a scalar quantity.
Power = Work done/ Time taken = ΔE/ Δt
Power = ΔPE/ Δt = Δmgh/ Δt
where, ΔPE is the change in potential energy,
Δh is the change in height,
Δt is the change in time,
ΔE is the change in energy,
Δh = 5.7 m,
Δt = 5.5 s,
m = 117 lb = 53.0703 kg
P = Δmgh/ Δt
P = (53 × 9.8 × 5.7)/ 5.5
P = (2960.58)/ 5.5
P = 538.28 watts.
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An artillery shell is fired with an initial velocity of 300 m/s at 55.5° above the horizontal. To clear an avalanche, it explodes on a mountainside 47.0 s after firing. What are the x- and y-coordinates of the shell where it explodes, relative to its firing point?
The x-coordinate of the shell where it explodes, relative to its firing point, is approximately 7688.1 meters, and the y-coordinate is approximately 9507.8 meters.
Is initial velocity always 0?If an automobile is at rest, its beginning velocity is 0.The starting velocity of an automobile that stops after using the brakes will be greater than zero, but the ultimate velocity will be zero.
The kinematic equations of motion can be used to resolve this issue.We can break down the initial velocity into its x- and y-components:
Vx = V0 cosθ = (300 m/s) cos(55.5°) ≈ 163.3 m/s
Vy = V0 sinθ = (300 m/s) sin(55.5°) ≈ 247.2 m/s
Next, we can use the equations of motion to find the shell's position after 47.0 seconds. The x-coordinate is given by:
x = Vx t = (163.3 m/s)(47.0 s) ≈ 7688.1 m
The y-coordinate is given by:
y = Vy t + (1/2)gt²= (247.2 m/s)(47.0 s) + (1/2)(9.81 m/s²)(47.0 s) ≈ 9507.8 m
where g is the acceleration due to gravity.
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Calculate the magnitude of the velocity of the rock just before it strikes the ground.
The magnitude of the velocity of the rock just before it strikes the ground is 25.48m/s.
Height = 17.0 m
Velocity = 33.0 m/s
Angle = 33.0°
[tex]-h= v_{y}t-\frac{1}{2}gt^2\\-17.0 = (33.0* sin33)t-4.9t^2\\4.9t^2-17.9t-17.0\\t= 4.435sec[/tex]
calculate the magnitude of the velocity of the rock just before it strikes the ground
[tex]v_y= (33.0*sin33)- 9.8*4.435\\v_y= -25.48m/s[/tex]
Velocity is a fundamental concept in physics that describes the rate at which an object changes its position in a given direction. It is a vector quantity that has both a magnitude and a direction, which means that it specifies not only how fast an object is moving, but also in which direction it is moving.
The formula for calculating velocity is velocity = displacement/time, where displacement is the change in the position of the object and time is the duration of that change. The SI unit of velocity is meters per second (m/s).
Velocity is an important concept in many areas of physics, including mechanics, thermodynamics, electromagnetism, and relativity. It is used to describe the motion of objects, the behavior of waves, and the interactions between particles. In addition to its use in physics, velocity is also a key concept in many other fields, such as engineering, astronomy, and sports science.
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Complete Question:
A man stands on the roof of a building of height 17.0 m and throws a rock with a velocity of magnitude 33.0 m/s at an angle of 33.0 ∘ above the horizontal. You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground.