Ask question

Ask question

All categories

2 years ago

14

A cart rolling at ground level has K = 1500 J. It comes to a ramp, rolls up the ramp to an elevated platform. When it's on this

elevated platform, it has K = 1000 J. What does its gravitational potential energy equal on top of that platform?

1 answer:

Black_prince [1.1K]2 years ago

8 0

The potential energy Wood be 1000

You might be interested in

What is the string theory.

Leno4ka [110]

String the·o·ry
noun
a cosmological theory based on the existence of cosmic strings.

7 0

2 years ago

Read 2 more answers

The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands

Ghella [55]

Answer:

Pressure on both feet will be $500N/m^2$

Explanation:

Weight of the person F = 500 N

Area of foot print $A=0.5m^2$

Area of both the foot $A=2\times 0.5=1m^2$

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure $P=\frac{F}{A}$

$P=\frac{500}{1}=500N/m^2$

So the pressure on both feet will be $500N/m^2$ when person stands on both feet.

7 0

2 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

2 years ago

Which of the following means that an<br> image is real?

Licemer1 [7]

Answer:_

Explanation:

7 0

2 years ago

Which ball has the greater average speed during the 1-s interval after release (assuming neither hits the ground during that tim

notka56 [123]

Answer:

The ball thrown downward

Explanation:

When the ball is thrown vertically, the acceleration of it is the gravity acceleration independent if it is thrown downward or upward. However, the acceleration is a vector, so, when the ball is thrown upward, the movement is against the gravity, so the acceleration is negative, and so, the velocity decreases during time; and when the ball is thrown downward, the movement goes to the gravity, so the acceleration is positive, so the velocity increase after time passes.

3 0

2 years ago