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Mamont248 [21]
3 years ago
9

An uncharged metal sphere, A, is on an insulating base. A second sphere, B, of the same size, shape, and material carrying charg

e Q is brought close to, but not touching, sphere A. Describe what happens to the charges on A and B as they are brought close but not touching. If we now remove sphere B and place it far away, what is the charge on sphere A
Physics
1 answer:
solong [7]3 years ago
3 0

Answer:

0

Explanation:

  • If we bring the charged sphere B close to, but not touching it , to the uncharged sphere A, as charges can move freely on the conductor, a charge -Q will be built on the outer surface of the sphere A, facing to sphere B.
  • As the sphere A must remain neutral, a charge Q will be built on the surface, on the side farther to the sphere B, as the following condition must be met:

       Q +(-Q) =0.

  • If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.
  • The total charge on sphere A must be 0, as there is no charge transfer from sphere B to sphere A.
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A 5.00-g bullet is fired into a 900-g block of wood suspended as a ballistic pendulum. The combined mass swings up to a height o
Thepotemich [5.8K]

Answer:

The value is  KE_b =0.710 \ J

Explanation:

From the question we are told that

   The mass  of the bullet is m_b  = 5.00 \ g  = 0.005 \  kg

  The mass  of the wood is  m_w =  900 \  g  =  0.90\  kg

   The height attained by the combined mass is  h =  8.0 \ cm  =  0.08 \ m

Generally according to the law of energy conservation    

    KE _b  =  PE_c

Here KE_b is the kinetic energy of the bullet before collision.

and PE_c is the  potential  energy of the combined mass of bullet and wood at the height h which is mathematically represented as

      PE_m  =  [m_b  + m_w] *  g *  h

So

   KE_b =PE_c   = [0.005  + 0.90] * 9.8 *0.08

=> KE_b =0.710 \ J

3 0
3 years ago
______ states that energy cannot be created or destroyed.
Pavel [41]
The answer is a newton second law
4 0
3 years ago
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A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto
sesenic [268]
<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float=l\times b\times h=5.2\times 2.4\times h=12.48h

Density of water=10^3kg/m^3

Weight of water displaced by barge=Buoyant force=-Weight of horse

Volume\;of\;water\times density\;of\;water\times g=410\times g

12.48h\times 1000=410

h=\frac{410}{12.48\times 1000}=0.03 m

1 m=100 cm

0.03 m=0.03\times 100=3cm

Hence, the depth of barge float=3 cm

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3 years ago
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What is the power of a refrigerator with voltage 110 V and<br> current 0.8 A?
tia_tia [17]

Answer: 88

Explanation:

8 0
3 years ago
A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

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