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zloy xaker [14]
2 years ago
12

A storm cloud has an electric charge of (2.1400x10^1) C near the top of the cloud and (2.99x10^1) C near the bottom of the cloud

. The magnitude of the electrostatic force between these two charges is (3.390x10^8) N. How tall is the cloud?​
Physics
1 answer:
blagie [28]2 years ago
4 0

Answer:

hey

Explanation:

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A Beam balance (or Beam scale) is a device to measure weight or mass. These are also known as mass scales, weight scales, mass balances, weight balances, or simply scales, balances, or balance scales. You can also use a graduated cylinder and determine the water displaced and figure out the mass of an object this way!

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3 0
3 years ago
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
14
Minchanka [31]

Answer:

C

Explanation:

8 0
2 years ago
Read 2 more answers
Which statement best describes the force of gravity
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Gravity<span> is measured by the acceleration that it gives to freely falling objects. At Earth's surface the acceleration of </span>gravity<span> is about 9.8 metres (32 feet) per second per second.</span>
4 0
3 years ago
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masya89 [10]

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

r=1.5(10)^{11} m is the orbital radius

a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}

a_{C}=0.0058 m/s^{2}

4 0
3 years ago
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