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kiruha [24]
4 years ago
12

In a thundercloud there may be an electric charge of 24 C near the top of the cloud and −24 C near the bottom of the cloud. If t

hese charges are separated by about 2 km, what is the magnitude of the electric force between these two sets of charges? The value of the electric force constant is 8.98755 × 109 N · m2 /C 2 .
Physics
1 answer:
lidiya [134]4 years ago
5 0

Answer:

Electric force, F=1.29\times 10^6\ N

Explanation:

Given that,

Charge 1, q_1=24\ C

Charge 2, q_2=-24\ C

Distance between charges, d = 2 km

The electric force is given by :

F=k\dfrac{q_1q_2}{d^2}

k is the electrostatic constant

F=8.98755\times 10^9\times \dfrac{(24)^2}{(2\times 10^3)^2}

F = 1294207.2 N

or

F=1.29\times 10^6\ N

Hence, this is the required solution.

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-4,200,000 J

Explanation:

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3 years ago
What are things measured by mass
Free_Kalibri [48]
Anything that is possibly solid or can be contained in a solid

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3 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m
forsale [732]

Answer:

Acceleration due to gravity will be g=5.718m/sec^2

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing =\frac{145}{100}=1.45sec

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

T=2\pi \sqrt{\frac{l}{g}}

So 1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}

\sqrt{\frac{0.55}{g}}=0.230

Squaring both side

{\frac{0.3025}{g}}=0.0529

g=5.718m/sec^2

So acceleration due to gravity will be g=5.718m/sec^2

3 0
4 years ago
Will mark the brainliest
AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

J = \Delta p = p_{f} - p_{i}

Where:                                                                

p_{f}: is the final momentum

p_{i}: is the initial momentum

The initial momentum is given by:

p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

v_{1_{i}} = -10 m/s

And because the nail is not moving:

v_{2_{i}}= 0                      

p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

Since the hammer and the nail are moving in the negative direction:

v_{1_{f}} = v_{2_{f}} = -9.7 m/s

p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s

Finally, the impulse is:

J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!                                                                                                                                                                                                                                                                                                                                                                                                                    

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3 years ago
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Your answer is correct: Resident 4 would provide the scientist with the most valid results.
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