State what is meant by a gravitational potential at point A is -1·70 × 109 J kg-1.

A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee

Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

$s = ut +\frac{1}{2}at^{2}$

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

$s = \frac{1}{2}gt'^{2}$

$s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}$ (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

$s = v\times t''$

$s = 343\times t''$ (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t' (3)

Using eqn (2) and (3):

s = 343(3.5 - t') (4)

from eqn (1) and (4):

$4.9t'^{2} = 343(3.5 - t')$

$4.9t'^{2} + 343t' - 1200.5 = 0$

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

3 0

2 years ago

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

5 0

2 years ago

A soccer player kicks a rock horizontally off a 37 m high cliff into a pool of water. If the player hears the sound of the splas

Alchen [17]

<h2>Answer:</h2>

(a). 16.741 m/s

(b). 15.75 m/s

<h2>Explanation:</h2>

In the question,

Height of the cliff, h = 37 m

Time taken to reach the sound to us = 2.92 s

Speed of the sound in air at room temperature = 343 m/s

Now,

Let us say the speed of the ball = u m/s

So,

Time taken by the ball to reach at the bottom, t is given by,

$t=\sqrt{\frac{2h}{g}}\\t=\sqrt{\frac{2\times 37}{9.8}}\\t=2.747\,s$

So,

Splash is heard after = 2.92 s

So,

Time taken by sound to travel the shortest distance along the hypotenuse of the triangle thus formed is,

t = 2.92 - 2.747

t = 0.172 s

Now,

Distance traveled by sound is given by,

$Distance=343\times 0.172\\Distance=59.02\,m$

So,

In the triangle using the Pythagoras Theorem,

Horizontal distance traveled is,

$D=\sqrt{59.02^{2}-37^{2}}\\D=45.99\,m$

So,

Speed of throwing of ball is given by,

$Distance=Speed\times Time\\45.99=u\times 2.747\\u=16.741\,m/s$

Therefore, the speed of the ball = 16.741 m/s.

(b).

If,

Speed of sound = 331 m/s

So,

Distance traveled by sound is,

$Distance=331\times 0.172\\Distance=56.932\,m$

So,

Distance traveled in the horizontal by ball is,

$Distance=\sqrt{56.932^{2}-37^{2}}\\Distance=43.269\,m$

So,

Speed of the ball thrown is given by,

$Speed=\frac{Distance}{Time}\\Speed=\frac{43.269}{2.747}\\Speed=15.75\,m/s$

Therefore, the speed of the ball = 15.75 m/s.

8 0

2 years ago

An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2?

Zigmanuir [339]

Force can be expressed as the product of mass and acceleration. Mathematically, that's F = m(a). Plugging the given into the equation, we have F = (13.5 kg)(9.5 m/s²) = 128.3 kg.m/s² or 128.3 N.

4 0

2 years ago

Zero, a hypothetical planet, has a mass of 5.8 x 1023 kg, a radius of 2.7 x 106 m, and no atmosphere. a 10 kg space probe is to

Tom [10]

To answer these questions just use the equations for potential energy using the mass and heights described. the potential energy at the prescribed heights = the initial kinetic energy required to reach that height.

Make sure you calculate the force of gravity on the surface using the radius of the planet.

5 0

2 years ago

State what is meant by a gravitational potential at point A is -1·70 × 109 J kg-1.​

State what is meant by a gravitational potential at point A is -1·70 × 109 J kg-1.