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sdas [7]
3 years ago
13

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

s 9.8 m/s 2 . 2 What is the gravitational potential energy associated with the ball relative to the ceiling? Answer in units of J. 012 (part 2 of 3) What is its gravitational potential energy relative to the floor? Answer in units of J. 013 (part 3 of 3) What is its gravitational potential energy relative to a point at the same elevation as the ball? Answer in units of J.
Physics
1 answer:
nydimaria [60]3 years ago
4 0

1. -23.2 J

The gravitational potential energy of the ball is given by

U=mgh

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

U=mgh

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J

3. 0 J

As before, the gravitational potential energy of the ball is given by

U=mgh

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J

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One mole of water is equivalent to 18 grams of water. A glass of water has a mass of 200 g. How many moles of water is this?
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Suppose a firm is producing 2,475 units of output by hiring 50 workers (W = $20 per hour) and 25 units of capital (R = $10 per h
Neko [114]

Answer

given,

firm is producing  = 2,475 units

output by hiring 50 workers W = $20 per hour

25 units of capital R = $10 per hour

marginal product of labor = 40

marginal product of capital = 25

\dfrac{MP_l}{MP_c}=\dfrac{40}{25}

\dfrac{MP_l}{MP_c}=\dfrac{8}{5}

\dfrac{W}{R}=\dfrac{20}{10}

\dfrac{W}{R}=2

\dfrac{MP_l}{MP_c} < \dfrac{W}{R}

Firm is not minimizing the cost because the firm use more capital and less labor.

3 0
3 years ago
A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
3 years ago
HELP URGENT- will give brainliest if correct
fredd [130]

Answer:

6 km is the right answer

hope it helps you

Explanation:

please mark me as brainliest

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