PHYSICS COLLEGE

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake

Answers

Answer 1

Answer:

Answer:

$|D_(depth) |=19.697m$

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

$x_(f)-x_(i)=v_(i)t+(1/2)at^(2)\n -5.0m=(o)t+(1/2)(-9.8m/s^(2) )t^(2)\n -4.9t^(2)=-5.0\n t^(2)=5/4.9\nt=√(1.02) \nt=1.01s$

As we have find the time taken now we need to find the final velocity vf from below equation as

$v_(f)=v_(i)+at\nv_(f)=0+(-9.8m/s^(2) )(1.01s) \nv_(f)=-9.898m/s$

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

$|D_(depth) |=|vt|\n|D_(depth) |=|(-9.898m/s)(1.99s)|\n|D_(depth) |=19.697m$

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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

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A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° above the horizontal. the coefficient of kinetic friction between the box and the surface is 0.350. what is the acceleration of the box?

Answers

The acceleration of the box is 0.81 m/sec².

What is acceleration?

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration

According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:

$\sum F= ma$ (1)

we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:

- the horizontal component of the force exerted by the rope, which is equal to

$F_x = F cos\theta = 250*cos 32 = 212 N$

the frictional force, acting in the opposite direction, which is equal to

$F_f = \mu *mg = 171.7 N$

By applying Newton's law (1), we can calculate the acceleration of the

box,

$F_x - F_f = ma\na = 0.81 m/sec^2$

The acceleration of the box is 0.81 m/sec².

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The acrylic plastic rod is 20 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Eₚ = 2.70 GPa, vₚ = 0.4.

Answers

Given Information:

diameter = d = 15 mm

Length = L = 20 mm

Axial load = P = 300 N

Eₚ = 2.70x10⁹ Pa

vₚ = 0.4

Required Information:

Change in length = ?

Change in diameter = ?

Answer:

Change in length = 0.01257 mm

Change in diameter = -0.003772 mm

Explanation:

Stress is given by

σ = P/A

Where P is axial load and A is the area of the cross-section

A = 0.25πd²

A = 0.25π(0.015)²

A = 0.000176 m²

σ = 300/0.000176

σ = 1697792.8 Pa

The longitudinal stress is given by

εlong = σ/Eₚ

εlong = 1697792.8/2.70x10⁹

εlong = 0.0006288 mm/mm

The change in length can be found by using

δ = εlong*L

δ = 0.0006288*20

δ = 0.01257 mm

The lateral stress is given by

εlat = -vₚ*εlong

εlat = -0.4*0.0006288

εlat = -0.0002515 mm/mm

The change in diameter can be found by using

Δd = εlat*d

Δd = -0.0002515*15

Δd = -0.003772 mm

Therefore, the change in length is 0.01257 mm and the change in diameter is -0.003772 mm

The data listed below are for mechanical waves. Which wave has the greatest energy?

Answers

theres no data listed below

Answer:

amplitude = 14 cm; wavelength = 7 cm; period = 12 seconds

Explanation:

1. Explain the change of state from solid dry ice to carbon dioxide gas.2. The motion of the particles in dry ice and carbon dioxide gas.

3. Explain how the original mass of dry ice compares with the mass of carbon dioxide gas.

Answers

1. Since the solid carbon dioxide never become liquid on melting under normal pressure. Thus through the process of sublimation, the solid carbon dioxide changes to gas

2. The molecules in dry ice, are in caged like structure just as the normal water ice but as it melts, the CO2 molecules having high affinity for gaseous state converts into a gas.

3. Dry ice is heavier than its gaseous form. Density of dry ice = 97.6 lb/cu.ft.

Density of carbon dioxide gas = 0.1144 lb/cu.ft.

With higher density and a fixed volume, mass of dry ice is higher than the CO2 gas

For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8 cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .

Answers

Answer:

Velocity = v = 2.24m/s

Acceleration = a = 0.20m/s²

Explanation:

Please see attachment below.

Given

z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

Final answer:

The roller coaster's velocity and acceleration at t=4 seconds is 7.64 m/s and 0.57 m/s² respectively.

Explanation:

The question is about understanding kinematics in cylindrical coordinates to analyze the motion of a roller coaster car. First, we need to understand that in polar coordinates, θ is changing with time t. So, the velocity vector v will have two components, one in the θ direction (rθ') and another in the z direction (z'). Given θ = 0.3t, we differentiate θ with respect to time to get θ' or dθ/dt = 0.3 rad/sec. Then, the z component of the velocity can be calculated by differentiating the equation of motion in the z-direction, z = -8 cos(θ), with respect to time. This gives z' = 8(0.3)sin(0.3t). So, at t=4s, z' = 8(0.3)sin(1.2) = 1.89 m/s. We then calculate rθ' = r*dθ/dt = 25*0.3 = 7.5 m/s.

The magnitude of velocity can then be calculated using the Pythagorean theorem: √((rθ')² + (z')²) = √((7.5)² + (1.89)²) = 7.64 m/s .

In a similar way, we can find the acceleration components. Given that r=25 m and is constant, radial acceleration is zero ( ar = r*(θ')²). The tangential acceleration is at = r*θ'' = r*d²θ/dt² =0 m/s² and z'' = dz'/dt = 8*0.3²*cos(0.3t). So, at t = 4s, z'' = 8(0.09)cos(1.2) = 0.57 m/s². The magnitude of the acceleration is given by √((ar)² + (at)² +(z'')²) = √((0)² + (0)² +(0.57)²)= 0.57 m/s².

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