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3 years ago

Explain why your body feels like it is being pushed back when the car starts back up again

Physics

2 answers:

Masteriza [31]3 years ago

8 0

<h2>Answer:</h2>

<u>This push is felt due to inertia</u>

<h2>Explanation:</h2>

Inertia is the resistance, of any physical object, to any change in its velocity. This property is also called the first law of motion which says a body in rest will remain at rest and a body in motion will stay in motion until an external force acts upon them. The same thing happens when our body is at rest and the car suddenly starts backing up again. Because the rest state of our body experiences a sudden push and our body resist it due to inertia.

Fittoniya [83]3 years ago

3 0

Answer:

Due to inertia.

Explanation:

As per law of inertia, a body will remain in its state of motion until an external force, unbalanced in nature, is applied on it. When the car is at rest, the person sitting in it will also be at rest. When the car suddenly moves, the body resist the motion due to inertia and it stays in initial position while the car moves. Due to this the body will be pushed back.

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Answer:

Explanation:

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Ps- The object will stay moving in the same speed and direction.

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2 years ago

An alpha particle (Î±), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elec

vaieri [72.5K]

Answer:

Speed = 575 m/s

Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

Explanation:

Given :

Potential difference, U = $-3.45 \times 10^{-3} \ V$

Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

So the potential difference for the alpha particle when it is accelerated through the potential difference is

$U=\Delta Vq_{\alpha}$

And the kinetic energy gained by the alpha particle is

$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2$

From the law of conservation of energy, we get

$K.E. = U$

$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

$v_{\alpha} \approx 575 \ m/s$

The mechanical energy is conserved in the presence of the following conservative forces :

-- electrostatic forces

-- magnetic forces

-- gravitational forces

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3 years ago

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Answer:

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3 years ago

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Answer:

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3 years ago

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marusya05 [52]

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