Which statement best compares potential and kinetic energy?

Physics

1 answer:

DaniilM [7]3 years ago

4 0

Answer: Kinetic energy increases and potential energy decreases when the velocity of an object increases

Explanation: probably that one

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kenny6666 [7]

Answer:

Explanation:

8 0

3 years ago

Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s

jok3333 [9.3K]

The initial height of the first body is given by:
$h_1 = \frac{1}{2}gt^2$
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
$h_1 = \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m$

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
$h_2 = \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m$

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.

6 0

3 years ago

Number of atoms of each element potassium chlorate?

11111nata11111 [884]

5 atoms is the answer

6 0

3 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$