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algol13
3 years ago
6

4.

Physics
1 answer:
marysya [2.9K]3 years ago
7 0
A. Terminal velocity
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Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.3×109 V and the quantity of charg
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Answer:

A) ΔU = 3.9 × 10^(10) J

B) v = 8420.75 m/s

Explanation:

We are given;

Potential Difference; V = 1.3 × 10^(9) V

Charge; Q = 30 C

A) Formula for change in energy of transferred charge is given as;

ΔU = QV

Plugging in the relevant values gives;

ΔU = 30 × 1.3 × 10^(9)

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B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.

This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.

Thus;

P.E = K.E

ΔU = ½mv²

Where v is final velocity.

Plugging in the relevant values;

3.9 × 10^(10) = ½ × 1100 × v²

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Read 2 more answers
When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and a
ryzh [129]

Complete question:

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?

Answer:

The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

Explanation:

Given;

combined mass of the shotgun and arm–shoulder, m₁ = 15 kg

mass of the projectile, m₂ = 0.04 kg

speed of the projectile, u₂ = 380 m/s

let the recoil velocity of the shotgun and arm–shoulder combination = u₁

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂ = 0

m₁u₁ = - m₂u₂

u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

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