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eduard
3 years ago
13

a thundercloud whose base is 500m above the ground. The potential difference between the base of the cloud and the m ground is 2

00MV. A raindrop with a charge of 4.0 x 10–12C is in the region between the cloud and the ground. What is the electrical force on the raindrop? A 1.6 x 10–6N B 8.0 x 10–4N C 1.6 x 10–3N D 0.40N
Physics
1 answer:
kolezko [41]3 years ago
3 0

Answer:

1.6×10⁻⁶ N.

Explanation:

From the question,

F = (V/r)q......................... Equation 1

Where F = Electric force on the raindrop, V = Potential difference between the base of the cloud and the ground, r = distance between the base of the cloud and the ground, q = the charge on a rain drop.

Given: V = 200MV = 200×10⁶ V, r = 500 m, q = 4.0×10⁻¹² C.

Substitute these values into equation 1

F = [(200×10⁶ )/500]×4.0×10⁻¹²

F = 1.6×10⁻⁶ N.

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A 7.91g bullet is fired into a 1.52-kg ballistic pendulum initially at rest and becomes embedded in it. The pendulum subsequentl
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Answer:

D

Explanation:

By law of conservation of momentum:

momentum before collision = momentum after collision

m_{1} u_1 + m_2 u_2  = m_1 v_1 + m_2  v_2\\ m_1 = mass of bullet\\u_1 = initial speed of bullet\\m_2 = mass of pendulum\\u_2 = initial speed of pendulum\\v_1 = final speed of bullet\\v_2 = final speed of pendulum

Initial speed of bullet is unknown whereas initial speed of pendulum will be zero as it was at rest.

Final speed of bullet and pendulum will be equal as bullet is embedded in pendulum and both moves together a vertical distance of 6.89cm.

Using third equation of motion:

2 a S = v_{f}^2  - v_{i}^2

where:

a = -g = acceleration due to gravity = 9.8 m/s^2 ( for vertical motion)\\S = h = 6.89cm = Vertical height or distance it covers\\v_f = final speed = 0 (after covering 6.89cm)\\v_i = initial speed (unknown)

Thus by placing values v_i = 1.1620 m/s

this speed will be final speed of collision for the calculation of initial speed of bullet.

Putting values:

(7.91) * x + (1.52*10^3) * 0 = (7.91 + 1.52*10^3) * 1.162\\7.91 x = 1775.431\\x = 224m/s

This 224m/s = 0.224Km/s which is closest to D

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Which statement would most likely be found in an advertisement from a <br> cell phone provider
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4 years ago
(25%) Problem 3: A bullet is fired horizontally into an initially stationary block of wood suspended by a string and remains emb
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Answer:

Thus, the initial velocity of the bullet is 507.5 m/s.

Explanation:

mass of bullet, m = 0.0085 kg

mass of block, M = 0.99 kg

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Let the initial velocity of bullet is u and the final velocity of block and bullet system is v.

Use conservation of energy

Potential energy of bullet bock system = kinetic energy of bullet block system

(M+m)\times g\times h = \frac{1}{2}\times \left ( M+m \right )v^{2}

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 0.95}=4.32 m/s

Now use conservation of linear momentum

mu = (M+m) v

0.0085 x u = (0.99 + 0.0085) x 4.32

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