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3 years ago

What is the equation for potential energy

Physics

1 answer:

meriva3 years ago

3 0

PE = mgh
where
m = mass
g = acceleration due to gravity
h = height

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Identify the energy transformation in the image

Yuliya22 [10]

Answer:

d one is correct

Explanation:

as the electrical energy in the socket is transferred to the electric tea pot

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2 years ago

what is the closest to the order of magnitude of the energy rquired to split a nucleus in to its individual protons and neutrons

skad [1K]

Answer:

More than two million electron volts.

Explanation:

More than two million electron volts energy are needed to break or split a deuteron into a proton and a neutron. Nuclear binding energy is the type of energy that is required to split an atom's nucleus into protons and neutrons. The deuteron is an isotope of hydrogen that is composed of a proton and a neutron and it is a stable particle. Very huge amount of energy is needed for the splitting of nucleus due to the presence of heavy particles i.e. proton and neutron.

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3 years ago

A race car decelerating by 3 m /s to 10 m / s has an acceleration of 25

Scilla [17]

Answer:

then it will stop

Explanation:

7 0

3 years ago

What is the effect in the picture

stepan [7]

The effect of this problem is that negative particles and positive particles contract to each other caused by electrical force.

8 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago