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3 years ago

Pls give me an example of newton's 1 ,2 ,3 law of motion pls this is due tomorrow

Physics

1 answer:

Bond [772]3 years ago

5 0

Answer:

I am not sure if this is a good time to get the latest flash player is required for video playback is unavailable right now because this video is not available for remote playback is unavailable right now because this video is

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Find relation for pressure at a depth h in a liquid of density d

klemol [59]

Can you elaborate more

5 0

3 years ago

Read 2 more answers

An oceanic ridge might be compared to what continental landform? A) delta B) great plains C) mountain range Eliminate D) mountai

Pavlova-9 [17]

A mountain range because an ocean ridge is an underwater mountain hope this helps you

6 0

3 years ago

Which of these terms means an object's resistance to changing its motion?

Mila [183]

Answer:

Inertia

F=ma

Action, reaction

All of the above

A heavy object requires more force to push than a lighter object.

8 0

3 years ago

A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What

Oksanka [162]

Answer:

The kinetic energy of the anti proton is 147.4 MeV.

Explanation:

Given that,

Energy = 2.12 GeV

Kinetic energy = 96.0 MeV

We need to calculate the kinetic energy of the anti proton

Using formula of energy

$E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}$

We know that,

$m_{p}c^2=m_{np}c^2$

So, $E_{photon}=2mc^2+K.E_{p}+K.E_{np}$

$K.E_{np}=E_{photon}-(2mc^2+K.E_{p})$

Put the value into the formula

$K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}$

$K.E_{np}=147.4\ MeV$

Hence, The kinetic energy of the anti proton is 147.4 MeV.

6 0

3 years ago

As a box slides down a ramp, friction does 23.0 joules of work. At the bottom of the ramp, the box has 3.8 joules of kinetic ene

tensa zangetsu [6.8K]

Answer:

The high of the ramp is 2.81[m]

Explanation:

This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.

If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.

We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.

$E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]$

Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.

$E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\$

And therefore

$m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]$

8 0

3 years ago