A 25 kg child is sitting at the top of a 4 m tall slide, what is his potential energy?

A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in

Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

$\frac{GMm}{d^{2}}=\frac{mv^2}{d}$

$v=\sqrt{\frac{GM}{d}}$

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

$v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}$

v = 1.81 x 10^-4 m/s

7 0

3 years ago

In Milgram's experiment:

SpyIntel [72]

Answer:

B. The "Learner" was working with Milgram.

Explanation:

just took the test

give brainliest, please. :)

3 0

2 years ago

A ball is released from rest at a height of 10 m and falls freely to the

ElenaW [278]

Answer:

The new kinetic energy would be 16 times greater than before.

Explanation:

Kinetic energy is found using this formula:

KE = 1/2mv²
where KE = kinetic energy (J), m = mass (kg), and v = velocity (m/s)

We can see that kinetic energy is directly proportional to the square of the velocity, meaning that if the speed was increased by 4 times, then the kinetic energy would get increased by a factor of 16.

The velocity just before the ball hits the ground can be found by the equation:

√(2gh)

Let's substitute h = 10 m and h = 40 m into this formula.

√(2g(10))
√(2g(40))

We can see that the velocity increases by a factor of 4 (10 m → 40 m).

Therefore, this means that the kinetic energy would also be increased by a factor of (4)² = 16. Thus, the answer is D) The new kinetic energy would be 16 times greater than before.

6 0

2 years ago

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th

Evgen [1.6K]

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

<u>We consider merry go-round as a ring:</u>

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

<u>Moment of inertia of the person considering as a point mass:</u>

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

<u>Now according to the conservation of angular momentum:</u>

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$

4 0

3 years ago

Read 2 more answers

Olivia is on a swing at the playground. at which point is her kinetic energy increasing and her potential energy decreasing? w x

Eva8 [605]

<h2>Olivia is on a Swing at Playground - Option 2 </h2>

Olivia is on a swing at the playground. Her kinetic energy increasing at x and her potential energy decreasing at x. At mean position velocity is maximum so kinetic energy ( K.E ) is also maximum and at mean position potential energy is minimum. Therefore, kinetic energy is increasing and potential energy decreasing at x.

5 0

3 years ago

Read 2 more answers