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Leni [432]
2 years ago
12

In the figure , if Q = 30 uC , q = 5.0 uC , and d = 30 cm , what is the magnitude of the electrostatic force on g?

Physics
1 answer:
lana [24]2 years ago
3 0

Answer:

F =  k\frac{q_1q_1}{ {d}^{2} }  \\  = 9.0 \times  {10}^{9}  (N {m}^{2}  {C}^{ - 2}) \frac{30 \times  {10}^{ - 6} (C)5.0 \times {10}^{ - 6} (C)}{ {0.3}^{2} } N

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An electron in a mercury atom changes from energy level b to a higher energy level when the
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3 years ago
An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the
poizon [28]

Answer:

The value is A  =  0.014 \  m

Explanation:

From the question we are told that

    The mass of the object is  m  =  2.0 \  kg

    The unstressed length of the string is  l  =  0.08 \  m

    The length of the spring when it is  at equilibrium is  l_e = 5.9 \  cm  =  0.059 \  m

      The initial speed (maximum speed)of the spring when given a downward blow v  =  0.30 \  m/s

Generally the maximum speed  of the spring  is mathematically represented as

           u =  A *  w

Here A is maximum height above the floor (i.e the maximum amplitude)

            and w is the angular frequency which is mathematically represented as

       w = \sqrt{\frac{k}{m} }

So

        u =  A *   \sqrt{\frac{k}{m} }

=>      A  =  u *   \sqrt{\frac{m}{k} }

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

           b  =  l -l_e

=>         b  = 0.08 - 0.05 9

=>         b  = 0.021 \  m

Generally at equilibrium position the net force acting on the spring is  

            k *  b  -  mg  =  0

=>         k *  0.021   -   2 * 9.8  =  0

=>        k =  933 \  N/m

So

            A  =  0.30  *   \sqrt{\frac{2}{933} }

=>          A  =  0.014 \  m

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3 years ago
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