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castortr0y [4]
3 years ago
12

What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 2.1 g that has been charged to -

3.0 nc ?
Physics
1 answer:
Liula [17]3 years ago
4 0

Answer:

Electric field, E=6.86\times 10^6\ N/C

Explanation:

It is given that,

Mass of sphere, m = 2.1 g = 0.0021 kg

Charge, q=-3\ nC=-3\times 10^{-9}\ C

We need to find the magnitude of electric field that balances the weight of a plastic spheres. So,

ma=qE

a = g

E=\dfrac{mg}{q}

E=\dfrac{0.0021\ kg\times 9.8\ m/s^2}{-3\times 10^{-9}\ C}

E=6860000\ N/C

or

E=6.86\times 10^6\ N/C

Hence, the magnitude of electric field that balances its weight is 6.86\times 10^6\ N/C. Hence, this is the required solution.

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An object falls from the top of a building that is 25 m high. Air resistance is negligible.
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The velocity of the object s calculated as 22.1 m/s.

<h3>What is the speed of the object?</h3>

Given that we can write that;

v^2 = u^2 + 2gh

Now u = 0 m/s because the object was dropped from a height

v^2 = 2gh

v = √2 * 9.8 * 25

v = 22.1 m/s

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Suppose you exert a 25-N force to lift a ball 0.4 m in 2 s. How much work is done?
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work is force x distance = 25 x 0.4

= 2.5x4 = 10joules

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Answer:

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(r= -82) is a example of what correlation
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A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert
ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

F_y=F-mg=8.5-0.46\times 9.8=3.992\ N

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2

Vertical displacement of rocket is same as the height of loop. So, y=20\ m

There is no initial velocity in the vertical direction. So, u_y=0\ m/s

Now, applying equation of motion in vertical direction. we have:

y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0
3 years ago
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