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sergejj [24]
3 years ago
15

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels

Physics
1 answer:
Aleksandr [31]3 years ago
3 0

Answer:

<em>(a) The average velocity is 16 m/s</em>

<em>(b) The acceleration is 0.4 m/s^2</em>

<em>(c) The final velocity is 24 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

v_f=v_o+at\qquad\qquad [1]

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

(c) From [2] we can solve for a:

\displaystyle a= 2\frac{x-v_ot}{t^2}

Since vo=8 m/s, x=640 m, t=40 s:

\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

v_f=8+0.4\cdot 40

v_f=8+16=24

The final velocity is 24 m/s

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What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?
sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

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F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F

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4 0
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Answer:

B. Containing charged regions

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So in the given case, the option b would be contributed to the molecules that have intermolecular forces

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8 0
2 years ago
3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?
muminat

Answer:

7 meters, 2.8 meters

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work done (nm) = force (n) * distance (m)

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140/50 = m

m= 2.8 meters

4 0
2 years ago
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