A boy jumps from a wall 3m high. What is an estimate of the change in momentum of the boy when he lands without rebounding?

A 5 kg ball is sitting on top of a hill. It has a Potential Energy of 6000J. What is the height of the ball?

natali 33 [55]

Explanation:

mass=5 kg

potential energy=6000j

height=?

Now

potential energy =m.g.h

or 6000=5*9.8*h

or 6000=49h

or 6000÷49=h

or h= 122.45m

6 0

3 years ago

Does the Earth stay in one place throughout the year? If it doesn't, describe its motion and where it's located in the solar syt

stiv31 [10]

the earth moves throughout the year such as rotate around the sun, so yes the it does move and it sits roughly at 93.048 million miles away from the sun. I hope this helps you out! :)

6 0

3 years ago

A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?

VLD [36.1K]

The initial kinetic energy of the car is
$E_1 = \frac{1}{2}mv_1^2 = 1.9 \cdot 10^3 J$

Then, the velocity of the car is decreased by half: $v_2 = \frac{v_1}{2}$
so, the new kinetic energy is
$E_2 = \frac{1}{2}mv_2 ^2 = \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}$
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
$E_2 = \frac{1.9 \cdot 10^3 J}{4}=475 J$

5 0

4 years ago

What makes up a atom

Licemer1 [7]

Answer:

They're typically made up of three main parts: protons, neutrons and electrons. Think of the protons and neutrons as together forming a “sun”, or nucleus, at the centre of the system. The electrons orbit this nucleus, like planets. If atoms are impossibly small, these subatomic particles are even more so.

Explanation:

hope i helped.

6 0

3 years ago

Read 2 more answers

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$