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3 years ago

What is the potential energy of a 3 kg rock sitting at the top of a 21 meter high cliff?

Physics

1 answer:

Svetradugi [14.3K]3 years ago

3 0

Answer:

618.03 J

Explanation:

Potential energy can be calculate using the following formula.

PE=m*g*h

where m is the mass, g is the acceleration due to gravity and h is the height.

In this situation, we have a 3 kilogram rock on a 21 meter high cliff. Therefore, the mass is 3 kg and the height is 21 m.

On Earth, the acceleration due to gravity is 9.81 m/s^2.

m=3 kg

h=21 m

g=9.81 m/s^2

Substitute these values into the formula.

PE=3*9.81*21

Multiply all the numbers together

PE=63*9.81

PE=618.03

The potential energy is 618.03 kg m^2/s^2, or 618.03 Joules

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An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed

Tema [17]

Let the mass of 2500 kg car be $m_1$ and it's velocity be $v_1$ and the mass of 1500 kg car be $m_2$ and it's velocity be $v_2$ .

After the bumping the mass be M and it's velocity be V.

By law of conservation of momentum we have

$m_1v_1+m_2v_2 = MV$

2500 * 5 + 1500 * 1=4000 * V

V = 14000/4000 = 7/2 = 3.5 m/s

So the velocity of the two-car train = 3.5 m/s

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3 years ago

You are riding a bicycle. If you apply a forward force of 125 N, and you and

erica [24]

Answer:

1.52g

Explanation:

Given parameters:

Force = 125N

Mass combined = 82kg

Unknown:

Acceleration of the bicycle = ?

Solution:

From Newton second law of motion suggests that:

Force = mass x acceleration

Acceleration = $\frac{force}{mass}$ = $\frac{125}{82}$ = 1.52g

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3 years ago

When a burning stick of increase is moved fast in a circle a circle of red light is seen.

anzhelika [568]

Answer:

The impression of the image on the retina lasts for about 1/16th of a second after the removal of the object. If a burning stick of incense is revolved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Explanation:

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2 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

Which type of surface would most likely be the best reflector of electromagnetic energy?

Len [333]

<span>light colored and smooth surface would most likely be the best reflector of electromagnetic energy.Light, shiny surfaces are the best reflectors of radiation and they will allow the waves to reflect and bounce off rather than absorb. we can consider mirror as the example ,it will only reflect the light energy falling on them and it will not absorb. The darker coloured and rough surfaced substances will definitely absorb some amount of light falling on it. so light coloured smooth or shiny surfaced material would be the best reflector for electromagnetic energy.</span>

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2 years ago