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arlik [135]
3 years ago
15

Three identical metal spheres are on insulating stands. Sphere A is given an initial charge of q. Sphere A is then briefly touch

ed to Sphere B. Afterward, Sphere B is then briefly touched to Sphere C. What amount of charge does each sphere now have?
. A

Sphere A - 0.25 q

Sphere B - 0.5 q

Sphere C - 0.25 q


B.

Sphere A - 0.25 q

Sphere B - 0.5 q

Sphere C - 0.25 q


C.

Sphere A - 0.33 q

Sphere B - 0.33 q

Sphere C - 0.33 q


D.

Sphere A - 0.5 q

Sphere B - 0.25 q

Sphere C .025 q
Physics
1 answer:
REY [17]3 years ago
4 0

Answer:

Hi fernyestrealla How r u

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You drop a 0.375 kg ball from a height of 1.37 m. It hits the ground and bounces up again to a height of 0.67 m. How much energy
Radda [10]

2.57 joule energy lose in the bounce .

<u>Explanation</u>:

when ball is the height of 1.37 m from the ground  it has some gravitational potential energy with respect to hits the ground  

Formula for gravitational potential energy given by  

Potential Energy = mgh

Where ,

m = mass  

g = acceleration due to gravity  

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

Potential Energy = 0.375\times9.8\times1.37

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

Potential Energy = 0.375\times0.67\times9.8

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

6 0
3 years ago
Two parallel wires carry a current I in the same direction. Midway between these wires is a third wire, also parallel to the oth
Luda [366]

Answer:

Force is repulsive hence direction of force is away from wire

Explanation:

The first thing will be to draw a figure showing the condition,

Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.

4 0
3 years ago
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Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo
Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, \omega=0.47\ rev/s=2.95\ rad/s

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

I={m_1r^2}

I={52\times 3.9^2}

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

I_2=\dfrac{m_2r^2}{2}

I_2=\dfrac{118\times (3.9)^2}{2}

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

I=I_1+I_2

I=790.92+897.39

I = 1688.31 kg-m²

The angular momentum of the system is :

L=I\times \omega

L=1688.31 \times 2.95

L=4980.5\ kg-m^2/s

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

8 0
3 years ago
Two point charges are separated by a distance of 1 meter. How much would the force change if one charge was 4x larger?
jekas [21]

Answer:

the force would increase 4 times more

Explanation

more force results more mass or acceleration

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2 years ago
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