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Mashcka [7]
3 years ago
14

A conducting spherical shell with inner radius ‘a’ and outer radius ‘b’ has a positive charge Q located at its center. The total

charge on the shell is -4Q, and it is insulated from its surroundings.
(a) Derive expressions for the electric field magnitude in terms of the distance ‘r’ from the center for the regions: r , a; a < r < b; r >b.
(b) What is the surface charge density on the inner surface of the conducting shell?
(c) What is the surface charge density on the outer surface of the conducting shell?
(d) Draw a sketch showing electric field lines and the location of all charges.
(e) Draw a sketch of the electric field E as a function of
Physics
1 answer:
kifflom [539]3 years ago
4 0

Answer:deeeznuts

Explanation:

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An object is pulled with a 85 N force inclined at 27° along a horizontal surface ABC
OLEGan [10]

Answer:

m = 15.15 kg

Explanation:

Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:

Fx = ma

F Cosθ = ma

where,

F = Magnitude of Force = 85 N

θ = Angle with horizontal = 27°

m = mass of object = ?

a = acceleration of object = 5 m/s²

Therefore,

85 N Cos 27° = m(5 m/s²)

m = 75.73 N/5 m/s²

<u>m = 15.15 kg</u>

6 0
3 years ago
A ball is spun around in circular motion such that it completes 50 rotations in 25 s. What is the frequency of its rotation? 2.
Elis [28]

Answer:

1. f = 2 Hz

2. f = 0.011 Hz

3. f = 0.067 Hz

4. t = 4 s

Explanation:

1. The frequency of rotation is given by:

f = \frac{\omega}{2\pi}

Where:

ω: is the angular speed = 50 rotations (revolutions) in 25 s.

We need to convert the units of ω.

\omega = \frac{50 rev}{25 s}*\frac{2\pi rad}{1 rev} = 4\pi rad/s

Now, the frequency is:

f = \frac{4\pi rad/s}{2\pi} = 2 Hz

2. The frequency is:

We know:

5 laps = 5 revolutions

t: time = 450 s

f = \frac{\omega}{2\pi} = \frac{\frac{5 rev}{450 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.011 Hz    

3. The frequency of the pendulum is:

f = \frac{\omega}{2\pi} = \frac{\frac{1 rev}{15 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.067 Hz

4. We have:

θ: number of revolutions = 48 rev

f = 12 Hz

t =?

The time can be calculated as follows:

f = \frac{\omega}{2\pi} = \frac{\theta}{2\pi t}

t = \frac{\theta}{2\pi f} = \frac{48 rev*\frac{2\pi rad}{1 rev}}{2\pi*12 Hz} = 4 s

I hope it helps you!

4 0
3 years ago
The horizontal surface on which the 2.0 - kg block slides is frictionless. The speed of the block before it touches the spring i
vlabodo [156]

Answer:

0.918 seconds

Explanation:

M = 2kg

V = 9.0 m/s

g = 9.8 m/s

F = Mg cosΘ

Θ = 0°

But F = Ma

a = v / t

F = m*v / t

Mv / t = mg cos0

V / t = g

t = v / g

t = 9.0 / 9.8

t = 0.918s

5 0
3 years ago
A student tosses a ball horizontally from a balcony to a friend 3.8 meters down below them. How long does the ball take to reach
Vsevolod [243]

Answer:

The time it takes the ball to fall 3.8 meters to friend below is approximately 0.88 seconds

Explanation:

The height from which the student tosses the ball to a friend, h = 3.8 meters above the friend

The direction in which the student tosses the ball = The horizontal direction

Given that the ball is tossed in the horizontal direction, and not the vertical direction, the initial vertical component of the velocity of the ball = 0

The equation of the vertical motion of the ball can therefore, be represented by the free fall equation as follows;

h = 1/2 × g × t²

Where;

g = The acceleration due gravity of the ball = 9.81 m/s²

t = The time of motion to cover height, h

Then height is already given as h = 3.8 m

Substituting gives;

3.8 = 1/2 × 9.81 × t²

t² = 3.8/(1/2 × 9.81) ≈ 0.775 s²

∴ t = √0.775 ≈ 0.88 seconds

The time it takes the ball to fall 3.8 meters to friend below is t ≈ 0.88 seconds.

8 0
3 years ago
Gravity is a force that pulls an object down towards the Earth. Can we see the force of gravity?
Alexeev081 [22]

Answer:

We can't actually see the force of gravity BUT we can see its effect.

6 0
3 years ago
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