All categories

2 years ago

How high up is a 3.00 kg object that has 300.j of energy

Physics

2 answers:

loris [4]2 years ago

8 0

Answer:

10.20meters

Explanation:

insens350 [35]2 years ago

7 0

Answer:

10.2 m

Explanation:

The formula for potential energy is

PE=mgh

Let's plug in the given values into the equation

300 J = 3.00 kg*9.8m/s²*h

We're assuming that the object is on Earth and has a gravitational acceleration of 9.8m/s²

300J = 29.4 N * h

h = 10.2 m

You might be interested in

Which method of heat transfer takes place when particles of matter vibrate and collide with each other in direct contact

Cerrena [4.2K]

Conduction, hope this helps! :)

6 0

3 years ago

Read 2 more answers

A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia

Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0

3 years ago

If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite d

Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

$v_{i}=-44.5m/s\\v_{f}=55.5m/s$

Now we need to find the initial momentum

$P_{1}=m*v_{i}$

Substitute the given values

$P_{1}=(0.2kg)(-44.5m/s)\\P_{1}=-8.9kg.m/s$

Now for final momentum

$P_{2}=mv_{f}\\P_{2}=(0.2kg)(55.5m/s)\\P_{2}=11.1kg.m/s$

So the change in momentum is given as:

ΔP=P₂-P₁

$=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s$

ΔP=20 kg.m/s

3 0

4 years ago

What kind of front occurs when a cold air mass replaces a warm air mass?

Andreyy89

Answer:

Explained below:

Explanation:

When a cold air mass meets a warm air mass, a front is formed and if the cold air is replacing the warm air, it is known as a cold front. Cold fronts frequently cause thunderstorms or rain showers because they pressure the air in a steep upward direction at the front's edge. They are also responsible to bring the changes in atmospheric pressure and wind direction.

7 0

4 years ago

There is a decrease in power. It can be concluded that there has also been a change in: A)current or voltage.

liraira [26]

A current or voltage

5 0

3 years ago