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kondor19780726 [428]
3 years ago
13

As a person pushes a box across a floor, the energy from the person’s moving arm is transferred to the box, and the box and the

floor become warm. During this process, what happens to energy?
Physics
2 answers:
Galina-37 [17]3 years ago
7 0
Hello! The answer would be conserved.
I hope this helped:)
Mark me brainilest if you get the chance please!
Have a great
san4es73 [151]3 years ago
3 0

Answer:

conserved

Explanation:

During this process the energy is conserved

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If a rock were dropped from a building it would hit the ground with a certain kinetic energy and velocity. If it fell from a roo
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Answer:

no

Explanation:

because if you test it they hit the ground at the same time.

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Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s and a speed of 42.0 m/s.
ss7ja [257]

Answer:

Force, |F| = 2100 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s, \dfrac{m}{t}=50\ kg/s

Initial speed, v = 42 m/s

The momentum is reduced to zero, final speed, v = 0

The relation between the force and the momentum is given by :

F=\dfrac{p}{t}

F=\dfrac{mv}{t}

F=50\ kg/s\times 42\ m/s

|F| = 2100 N

So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.

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2 years ago
It is an employer's responsibility to provide training for special equipment. true false
Naddik [55]

Answer:

True I hope you like it

Give me a brainliest answer

6 0
2 years ago
The Sex Equity in Education Act requires educational institutions to distribute a sexual harassment policy to:
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4 0
2 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

Where in this case:

m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

Therefore:

E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

8 0
3 years ago
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