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4 years ago

Describe the relationship between a moving object's mass and its kinetic energy.

Physics

1 answer:

Paladinen [302]4 years ago

5 0

Answer:

$E=\dfrac{1}{2}mv^2$

Explanation:

If m is the mass of the object and v is the velocity of the object. The kinetic energy is due to the motion of an object. It is given by the relation as follows :

$E=\dfrac{1}{2}mv^2$

The above formula is used to find the kinetic energy of an object.

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A concrete block (B-36 x10 °C-') of volume 100 mat 40°C is cooled to

ruslelena [56]

T1=40°C=313K
T_2=-10°C=263K

Applying Charles law

$\\ \sf\Rrightarrow \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\\ \sf\Rrightarrow \dfrac{100}{313}=\dfrac{V_2}{263}$

$\\ \sf\Rrightarrow V_2=\dfrac{26300}{313}$

$\\ \sf\Rrightarrow V_2=84.02ml$

6 0

2 years ago

Based on the free-body diagram, the net force acting<br> on this firework is

Strike441 [17]

0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

Fnet = 9,452N - 9,452N = 0N

4 0

3 years ago

Read 2 more answers

Convert the arc length of 107869221.61 revolutions into rotation

Genrish500 [490]

Answer:

1 r = 6.283185 rad

677,762,308.31685 rad

Explanation:

8 0

3 years ago

What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$