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Kipish [7]
3 years ago
12

Negative charge –Q is distributed uniformly over the surface of a thin spherical insulating

Physics
1 answer:
loris [4]3 years ago
3 0

Answer:

Explanation:

The force of attraction or repulsion is given by

F=q1q2/4πϵ0r²

A. It is given that q1= -Q i.e a negative charge

And q2=q and it is at a distance r from Q.

Since it is a positive and negative charge then it is a force of attraction

So applying the equation,

F=q1q2/4πϵ0r²

F=Qq/4πϵ0r²

B. The direction of the force on q due to the Q is force of attraction and it is moving towards each other.

C. It is given that q1= -Q i.e a negative charge

And q2=q and it is at a distance r from Q.

Since it is a positive and negative charge then it is a force of attraction

So applying the equation,

F=q1q2/4πϵ0r²

F=Qq/4πϵ0r²

But if r is less than R i.e it is inside the sphere, then the F is zero inside the sphere

F=0N

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Explanation: An atom of potassium would have a very easy time losing one electron to form an ionic bond with an atom of an element that would easily accept it.

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How to find density of a gas.
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Subtract the mass of the cylinder from the mass of the cylinder when it contains gasoline. This is the mass of the gasoline. Divide this figure by the volume, 100 ml, to get the density.

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A bird has a mass of 0.8 kg and flies at a speed of 11.2 m/s. How much kinetic energy does the bird have?
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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i
drek231 [11]

Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = \frac{28}{10} × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

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t = 0.2845Nm (rounded to 4 decimal places)

8 0
3 years ago
A 7.25 kg7.25 kg block is sent up a ramp inclined at an angle ????=28.5°θ=28.5° from the horizontal. It is given an initial velo
Free_Kalibri [48]

Answer:

15.03 m

Explanation:

Given:

mass of the block, m = 7.25 kg

Angle, Θ = 28.5°

Initial speed of the block, v₀ = 15 m/s

let the distance traveled by the block be 's'

Now, applying the work energy theorem,

we have

(m\times g\times\sin(\theta)\times s) + \mu_k\times mg\times s\times cos(\theta) = \frac{1}{2}\times m\times v^2

on substituting the values in the above equation, we get

(7.25\times 9.8\times\sin(28.5^o)\times s) + 0.326\times 7.25\times9.8\times s\times cos(28.5^o) = \frac{1}{2}\times 7.25\times 15^2

or

33.902\times s) +20.35\times s = 815.625

or

54.252\times s = 815.625

s = 15.03 m

Hence, the block will travel 15.03 m up the ramp

6 0
3 years ago
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