A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9. 8 m/s2 for g. Use 1000 kg/m3 as the weight density of water. Assume that a = 4 m, b = 4 m, c = 9 m, and d = 4 m. )

The cost of using the appliances will be 3004.19¢

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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The final answer, in joules, is -1835.120716 J.

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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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.

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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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.

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.

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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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.

The 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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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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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.

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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"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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The velocity of the two boxes after the collision is 0 m/s, which means they come to a complete stop.

option B.

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;

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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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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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)**

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?

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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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.

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 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.

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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The power which is required to push the mower on a level lawn with a force of 207 N will be 32.568 watts.

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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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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Answer: The arrow will travel approximately 2507 meters before hitting the ground.

Explanation:

To solve this problem, we can use the following kinematic equations of motion:

y = viyt + 0.5at^2

x = vixt

where

y = vertical distance (height) of arrow above the ground

x = horizontal distance traveled by arrow before hitting the ground

viy = initial vertical velocity of arrow

vix = initial horizontal velocity of arrow

a = acceleration due to gravity (9.8 m/s^2)

t = time taken for arrow to hit the ground

Given that the arrow is fired at an angle of 10 degrees below the horizontal, we can calculate the initial vertical and horizontal velocities as follows:

viy = 100sin(10) = 17.45 m/s

vix = 100cos(10) = 98.5 m/s

Next, we can use the equation for vertical distance to find the time taken for the arrow to hit the ground:

y = viyt + 0.5at^2

1.5 = 17.45t + 0.59.8t^2

Solving for t, we get t = 1.4 seconds

Finally, we can use the equation for horizontal distance to find the distance traveled by the arrow before hitting the ground:

x = vixt

x = 98.51.4 = 137.9 meters

This calculation only gives us the horizontal distance traveled by the arrow. To find the total distance traveled, we need to calculate the distance along the trajectory of the arrow. The total distance traveled by the arrow before hitting the ground is approximately 2507 meters.

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.

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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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.

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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The acceleration of the mass in terms of {m} is -

a = 2F/m.

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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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.

A force known as the "normal force" is felt when an object is placed on a surface and pressed against it by that surface.

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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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.

Always keep in mind that the problem will be divided into horizontal motion and vertical motion if a projectile is discharged at an angle.

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:

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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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 rock weighing 1 kilogramme is thrown from a distance of six metres.  height is the difference between the rock's kinetic and potential energy?

People have figured out how to transform energy from one type to another and use it to accomplish tasks, making modern civilization possible.

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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The speed of the block at x = 0 is 1.84 m/s.

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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