All categories

2 years ago

If a 5 kg bowling ball is lifted to a height of 2 meters, how much potential

Physics

1 answer:

natka813 [3]2 years ago

8 0

Answer:

$\boxed {\boxed {\sf 98.1 \ Joules}}$

Explanation:

The formula for potential energy is:

$E_p=mgh$

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass is 5 kilograms and the height is 2 meters. Assuming this ball is on Earth, then the gravitational acceleration is 9.81 meters per square second.

$m= 5 \ kg \\g= 9.81 \ m/s^2 \\\h= 2 \ m$

Substitute the values into the formula.

$E_p= ( 5 \ kg )(9.81 \ m/s^2)( 2 \ m)$

Multiply the first two numbers.

$E_p=49.05 \ kg*m/s^2 (2 \ m )$

Multiply again.

$E_p= 98.1 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer is equal to 98.1 Joules

$E_p= 98.1 \ J$

The ball has 98.1 Joules of potential energy.

You might be interested in

As 390 g of hot milk cools in a mug it transfers 30 000 j of heat to the environment. whats is the temperature change of the mil

Fed [463]

You have to use the specific heat equation.

Q = cmΔT where Q is the energy, c is specific heat, m is mass, and ΔT is change in temp.

So we can substitute our variables into the equation.

30000J = (390g)(3.9J*g/C)ΔT

Solving for ΔT, we get:

30000J/[(390g)*(3.9J*g/C) = ΔT

ΔT = 19.72386588C

I'm assuming the temperature is C, since it was not specified.

Hope this helps!

8 0

3 years ago

For the sun, which is the final step in core fusion?

Marina86 [1]

The answer would be stage of<span> conversion of Hydrogen to Helium.
At the end of this phase, the helium content of the suns will have the popping effect where more and more helium is now on the core resulting to higher temp and density as helium converted from hydrogen is much heavier.</span>

3 0

2 years ago

Energy that flows from an objective with a higher temperature to an object with a lower temperature

dalvyx [7]

That's THERMAL energy, often referred to as "heat".

7 0

3 years ago

A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the

ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

$\frac {Vf^{2}-Vo^2}{2.a} =X$

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

$Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s$

the speed with the ball hits the ground is 32.46m/s

8 0

2 years ago

Read 2 more answers

A top fuel dragster covers the quarter mile in 4.5 seconds. What is it's speed in miles per hour

gladu [14]

<span>Distance in Miles (.25) x 3600 (seconds in an hour) / time in seconds = 200 MPH. Drag racing calculates the top speed via a speed trap starting 66 feet from the finish line.</span>

6 0

3 years ago

Read 2 more answers