Ask question

Ask question

All categories

3 years ago

6

A chain 72 meters long whose mass is 29 kilograms is hanging over the edge of a tall building and does not touch the ground. How

much work is required to lift the top 14 meters of the chain to the top of the building?
Use that the acceleration due to gravity is 9.8 meters per second squared. Your answer must include the correct units.

1 answer:

dangina [55]3 years ago

5 0

Answer:

Work done required is 3567.2 J

Explanation:

Given :

Length of chain, l = 72 m

Mass of chain, M = 29 kg

Linear mass density of chain, μ = $\frac{Mass\ of\ chain }{Length\ of\ chain}$ = $\frac{29}{72}$ = 0.40 kg/m

Let x be the length of the chain which lift to the top of the building.

Work done required to lift the chain is equal to the potential energy of the chain.

W = ∫μg (72 - x ) dx

Here g is acceleration due to gravity.

The limit of integration is from 0 to 14.

W = μg ( 72x - x²/2)

Substitute 0.40 kg/m for μ, 9.8 m/s² for g and 14 m for x in the above equation.

W = $0.40\times9.8\times(72\times14\ - \frac{14^{2} }{2})$

W = 3567.2 J

You might be interested in

A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?

fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as $V_1$ , the relation with the final volume as

$V_2 = V_1 +0.163\% V_1$

$V_2 = V_1 +0.00163V_1$

$V_2 = 1.00163V_1$

Initial temperature = $21.1\°C$

Let T be the temperature after expanding by the formula of volume expansion

we have,

$V_2 = V_1 (1+\gamma \Delta t)$

Where $\gamma$ is the volume coefficient of copper $5.1*10^{-5}/C$

$1.00163V_1 = V_1(1+\gamma(T-21.1\°))$

$1.00163 = 1+5.1*10^{-5}(T-21.1\°)$

$0.00163 = 0.000051T-0.0010761$

$T = 53.0608\°C$

Therefore the temperature is 53.06°C

7 0

3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

4 0

3 years ago

During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61

bonufazy [111]

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 = y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

7 0

3 years ago

What has been happening to the cosmic microwave background radiation since the Big Bang?

kondor19780726 [428]

Answer:

Explanation:

Cosmologists refer to a "surface of last scattering" when the CMB photons last hit matter; after that, the universe was too big. So when we map the CMB, we are looking back in time to 380,000 years after the Big Bang, just after the universe was opaque to radiation. But the CMB was first found by accident.

plz mark as brainliest

6 0

3 years ago

6 Fig. 6.1 is a full-scale diagram that represents a sound wave travelling in air

Oxana [17]

From the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

<h3>What is the frequency of a wave?</h3>

The frequency of a wave is the number of complete oscillation per second completed by a wave.

Frequency is related to wavelength and speed by the following formula:

Frequency = velocity/wavelength

Velocity of sound in air = 330 m/s

The measured wavelength = 5.0 cm = 0.05 m

Frequency = 330/0.05 = 6660 Hz

Therefore, based on the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

Learn more about frequency of sound at: https://brainly.in/question/15373132
#SPJ1

7 0

1 year ago