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4 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]4 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]4 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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An object is moving north with an initial velocity of 14 m/s accelerates 5m/s for 20 seconds. What is the final velocity of the

olga2289 [7]

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4 years ago

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

4 years ago

A 1.2kg stone is tied to a string and swung in a vertical circle with a radius of 0.75m. The string can withstand a tension of 4

san4es73 [151]

Hello!

We can begin by summing the forces acting on the stone when it is at the bottom of its trajectory.

Refer to the free-body diagram in the image below for clarification.

We have the force of tension (produced by the string) and the force of gravity acting in opposite directions, so:
$\Sigma F = T - F_g$

The net force is equivalent to the centripetal force experienced by the stone. Recall the equation for centripetal force for uniform circular motion:
$F_c = \frac{mv^2}{r}$

m = mass of object (1.2 kg)
v = velocity of object (? m/s)
r = radius of circle (0.75 m)

The centripetal force is the resultant of the forces of tension and gravity, and points upward (same direction as the tension force) since the tension force is greater.

Therefore:
$\frac{mv^2}{r} = T - Mg$

We can solve the equation for 'v':

$mv^2 = r(T - Mg) \\\\v^2 = \frac{r(T - Mg)}{m}\\\\v = \sqrt{\frac{r(T - Mg)}{m}}$

Plug in values and solve.

$v = \sqrt{\frac{(0.75)(40 - 1.2(9.8))}{1.2}} = \boxed{4.201 \frac{m}{s}}$

3 0

2 years ago