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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Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288 lb football player. << read less.
It would be easier to tackle the 220 lb player because less force is required to change their motion due to their lower inertia.
Force is the interaction between two objects that can cause a change in motion. Weight is a type of force that is caused by gravity acting on an object with mass. Mass is a measure of the amount of matter in an object, and it determines the strength of the gravitational force acting on the object. Inertia is the tendency of an object to resist changes in its state of motion.
When tackling a football player, the player exerts a force on the tackler, and the tackler exerts an equal and opposite force on the player due to Newton's third law of motion. The player's weight is also a force acting on the tackler, and it depends on the mass of the player and the strength of the gravitational field.
The mass of an object determines its inertia, which is the resistance of the object to changes in its state of motion. In other words, an object with a greater mass has a greater inertia, and it requires a greater force to change its motion. Therefore, when tackling a football player, it is easier to tackle a player with a lower mass, or in other words, a lower weight, than a player with a higher mass, or a higher weight.
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What is INTEGRITY? Why is it necessary in our school community? Why is it necessary in society in general? What does it look like in practice?
Integrity is the quality of being honest, truthful, and morally upright. It means doing the right thing even when no one is watching, and being accountable for one's actions.
Why is Integrity necessary?
In our school community, integrity is essential as it ensures that students and faculty members act with honesty and ethics in all their interactions. It creates a safe and trustworthy environment where individuals are respected, and their contributions are valued. Without integrity, academic performance, reputation, and character can be damaged, which can negatively impact the school community.
Integrity is also necessary in society in general, as it ensures that individuals act with honesty, ethics, and fairness in all their interactions. It helps to create a sense of trust and respect among people, which is essential for building positive relationships and fostering a healthy community. In business and professional settings, integrity is crucial as it helps to build credibility, establish trust, and maintain good relationships with clients and stakeholders.
In practice, integrity looks like individuals acting with honesty, ethics, and moral values in all their interactions. It means taking responsibility for one's actions, being accountable for one's mistakes, and working towards making things right when something goes wrong. It also means being respectful of others, treating them with dignity, and valuing their contributions. In the school community, it looks like students and faculty members conducting themselves with honesty and integrity in academic and personal settings.
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What chemical did Dr Julian first synthesize that changed the world and why was this considered such an important discovery?
Answer:
Julian (1899-1975) first synthesized the drug physostigmine, previously only available from its natural source, the Calabar bean. His pioneering research led to the process that made physostigmine readily available for the treatment of glaucoma.
Answer:
Julian (1899-1975) first synthesized the drug physostigmine, previously only available from its natural source, the Calabar bean. His pioneering research led to the process that made physostigmine readily available for the treatment of glaucoma.
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 each of the images, two blocks are attached to a balance. Use the position and size of the blocks to determine which block is more dense in each image.
A scale weighs two boxes. The green box on the left has a large volume and pulls lower on the scale. The blue box on the right has a smaller volume and is higher on the scale.
Which block is more dense in the first image?
Which block is more dense in the second image?
Which block is more dense in the third image?
First image: Green block is larger and heavier than the blue block.
Second image: Blue and green block are the same size, but blue is heavier
Third image: Green block is smaller and heavier than the blue block
It is impossible to estimate the relative density of the blocks. The rightmost blue chunk is denser. The chunk of green on the left is denser. The mass of a material per unit volume is the definition of density.
Describe density with an example:Its density refers to how much material can fit into a specific quantity of space. For instance, a block of lead (Pb), which is heavier and less durable, will have a lower density than a block of gold (Au), which is stronger and lighter (Au).
Mass and density: what are they?The term "mass" refers to a thing's total amount of matter. The term "density" describes how tightly atoms are packed or how close together they are in a material. To calculate inertia, utilize the idea of mass.
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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.
A pulley consists of a large disk of radius R and a small disk of radius r that are welded together and mounted on a horizontal frictionless axle through their common centers. The moment of inertia of this pulley is I. A box containing a turkey of mass M is suspended from a rope wound around the large disk; a box containing a pumpkin of mass m is suspended from a rope wound around the small disk. The ropes do not slip on the disks. The system is released from rest and the turkey begins to descend, while the pumpkin is being lifted up.
Use forces and torques to derive an expression for the angular acceleration of the pulley, in terms of system parameters
The angular acceleration of the pulley is given by the expression α = (MgR - mgr) / I.
The forces acting on the system are the gravitational forces on the turkey and the pumpkin, which are Mg and mg, respectively. T
These forces create torques on the pulley, with the torque on the large disk being MgR and the torque on the small disk being -mgr. The negative sign on the torque of the small disk indicates that it is acting in the opposite direction of the torque on the large disk.
The sum of these torques is equal to the moment of inertia of the pulley times its angular acceleration, which can be written as:
MgR - mgr = Iα
Solving for the angular acceleration, α, we get:
α = (MgR - mgr) / I
This is the expression for the angular acceleration of the pulley in terms of the system parameters. It shows that the angular acceleration depends on the masses of the turkey and the pumpkin, the radii of the large and small disks, and the moment of inertia of the pulley.
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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.
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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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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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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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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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 × 105 Pa. What force due to such an explosion will be exerted on the side of a house with dimensions 2.9 m × 11 m? Answer in units of N.
The force exerted on the side of the house is due to the given explosion.
From the formula of pressure,
Where,
- pressure = 0.2 atm × 1.013×10⁵ Pa =
- Force = ?
- area = 3.89 m × 22.6 m =
So,
Put the values in the formula,
Therefore, the force exerted on the side of the house is due to the given explosion.
What is determined by calculating the slope of the position vs time graph? position velocity distance displacemen
in an electroscope being charged by induction, what happens when the charging rod is moved away before the ground is removed from the knob?
The electroscope will remain charged and continue to show the presence of an electric charge if the ground is withdrawn from the knob before the charging rod is removed.
An early scientific tool used to find electrical charge on a body is called an electroscope. The movement of a test object caused by the Coulomb electrostatic force is used to detect charge. Voltage and charge on an object are inversely correlated.
There are three traditional electroscope types: the needle electroscope, the gold-leaf electroscope, and the pith-ball electroscope (third).
electroscopes are utilized to determine whether a body has an electric charge. The movement of a test object caused by the Coulomb electrostatic force is used to detect charge. Voltage and charge on an object are inversely correlated.
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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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When does magnetic force push objects apart?
A: when opposite poles of two magnets are facing each other
B: never
C: when like poles of two magnets are facing each other
D: when the objects are made of wood
When two magnets' opposing poles are lined up with one another (option A).
Is strength equivalent to force?As a result of their seeming comparable definitions, force and power are frequently employed interchangeably. They are not equivalent, nevertheless, in physics. Power is a statement of energy used over time, of which force is a component, while force is the essential outcome of a collision between two objects.
What generates force?Every time two things interact, a force is applied to each of them. The force is lost when the interaction between the two items ends. Only through interactions can forces be created.
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Compare the normal force that the seat exerts on you to the force that earth exerts on you when you are passing the bottom of the loop.
A) At the bottom, normal force is minimum so, Nbottom = mvb²/R - mg
B) At the top, normal force is maximum so, Ntop = mvt²/R + mg
A) When you are passing the bottom of the loop, the normal force that the seat exerts on you can be compared to the force that Earth exerts on you. At the bottom of the loop, the net force on you is equal to the force of gravity, and it acts towards the centre of the loop, providing the centripetal force required to keep you moving in a circular path.
The normal force, N, is the force exerted by the seat on you, perpendicular to the surface of the seat. Gravity's energy, mg, operates vertically downward. Therefore, we write,
Nbottom + mg = Fnet = mvb²/R
where,
R is radius of the loop
vb is the speed at the bottom of the loop
Nbottom = mvb²/R - mg
B) When you are passing the top of the loop, the normal force that the seat exerts on you can be compared to the force that Earth exerts on you. At the top of the loop, the net force on you is again equal to the force of gravity, but it acts downwards, providing the centripetal force required to keep you moving in a circular path.
The normal force, N, is still perpendicular to the surface of the seat. Gravity's energy, mg, operates vertically downward. Therefore, we can write:
The normal force is at its lowest point at the bottom of the loop and is equivalent to,
Ntop + mg = Fnet = mvt²/R
where,
vt is the speed at the top of the loop
At the top of the loop, the normal force is at a maximum, and it is equal to,
Ntop = mvt²/R + mg
The question is incomplete. The complete question is 'You ride a roller coaster with a loop-the-loop.A) Compare the normal force that the seat exerts on you to the force that Earth exerts on you when you are passing the bottom of the loop. Express your answer in terms of R (radius of the loop), vb (speed at the bottom of the loop), and constant g.B) Compare the normal force that the seat exerts on you to the force that Earth exerts on you when you are passing the top of the loop. Express your answer in terms of R (radius of the loop), vt (speed at the top of the loop), and constant g.'
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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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A fishing bont weighing 196,200 N is towed by a much larger ship. The tow cable in linearly elastic and elongates 0.03 m for each 500 N of tension in it. The wave and viscous drage on the fishing boat can be used to be linearly proportional to its velocity, and equal to 10,000 N/m. At time to the large tow ship starts moving with constant velocity, V = 3 m/s. There is no initial slack in the cable ) Propee a model of this system consisting of ideal lumped elements. (Mn, springs, dampers) Do this carefully; the rest of the problem depends on its accuracy. b) Find an expression for the fishing boat velocity was a function of time. Plot the velocity of both boat on the same graph c) What is the maximum force in the cable and at what time does it occur? d) What is the elongation of the tow cable due to the drag of the fishing boat at t=0 e) It is desired to change the stiffness of the cable to that the fishing boat will approach the velocity of the tow ship as fast as posible without cecillating What should the cable stiffness be in this case?
The spring, damper, and mass make up this system's model. The linear elasticity of the tow cable is represented by the spring.
The fishing boat's viscous drag and wave are represented by the damper. The fishing boat itself is represented by the mass.
b) The following expression can be used to represent the fishing boat's velocity over time:
V(t) = V₀ + [Ft - (10,000/m) ×Vt]/M,
where V₀ is the fishing boat's initial velocity, Ft is the cable force, and M is the boat's mass.
c) When the fishing boat's velocity equals the tow ship's velocity, the cable experiences its maximum force.
This occurs at t=0, and the force in the cable is 196,200 N.
d) The fishing boat's drag causes the tow cable to elongate by 0.03 m at t=0.
e) The stiffness of the cable should be 196,200 N/0.03 m to ensure that the fishing boat reaches the tow ship's velocity as quickly as possible without oscillating.
How does velocity work?The rate of change in an object's position in a particular direction is measured by its velocity. Because it is a vector quantity, it has a magnitude (also known as speed) and a direction. Using the derivative of the object's position over time, velocity can be determined. The calculation of a particle's or object's motion over time frequently makes use of velocity.
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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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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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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?
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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QUESTION 2
What are propagating in electromagnetic waves?
Electric field and magnetic field.
Electrons.
Positive charges.
Air.
Electromagnetic waves propagate through the interaction of : A. electric and magnetic fields.
What are propagating in electromagnetic waves?Electric and magnetic fields: As the electric field oscillates, it creates a changing magnetic field, and vice versa. These changing fields propagate through space, creating a self-sustaining wave that can travel through a vacuum or a medium such as air.
Electrons are not directly involved in the propagation of electromagnetic waves, but they can be affected by the waves as they interact with matter.
Positive charges also do not directly participate in the propagation of electromagnetic waves, but they can be influenced by the electric field component of the wave.
Air or any other medium does not play an active role in the propagation of electromagnetic waves, but it can affect the behavior of the waves.
Therefore the correct option is A.
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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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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)**
WORTH 40 POINTS!!
Label the following geologic map. Click on a label below the map to select it, and then click on the appropriate box on the map to place the label.
1.) Rock age/name
2.) Strike and dip
3.) Rock type
4.) Contact line
5.) Fault line
Geologic map: The correct label as placed in the respective boxes are as follows- is attached to the answer.
What is Geologic map?A geologic map is a scientific illustration that depicts the physical features of a geographic area and the underlying geology. It is used to help interpret the geologic history of a region. Geologic maps are created by geologists and contain information about the distribution of different rock types and other subsurface features. This information is used to interpret the area’s geologic history, understand the resources available in the area, and plan for development and management of the area. Geologic maps include information such as the age and type of rocks, their structures and minerals, faults and folds, and the presence of any surface features such as rivers, lakes, and soils. Geologic maps are used by geologists, engineers, planners, and other professionals to help understand the underlying geology and make decisions about how to use the land.
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In a 42 s interval, 597 hailstones strike a glass window of area 0.597 m2 at an angle 50◦ to the window surface. Each hailstone has a mass of 6 g and speed of 6.3 m/s. If the collisions are elastic, find the average force on the window.
Answer in units of N.
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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