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3 years ago

Please help me

Physics

1 answer:

Grace [21]3 years ago

6 0

Answer:

a. 60 N*s

b. 60 (kg*m)/s

c. 3 m/s

Explanation:

Givens:

m = 20 kg

v_i = 0 m/s

t = 10 s

F = 6 N

a) Impulse:

I = F*t

I = 6 N*10 s

I = 60 N*s

b) Momentum:

p = v*m

F = m(a)

a = F/m

a = 6 N/20 kg

a = 0.3m/s^2

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

p = 3 m/s*20 kg

p = 60 (kg*m)/s

c. Final velocity

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

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Answer:

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3 years ago

Consult Multiple-Concept Example 5 to review the concepts on which this problem depends. A light bulb emits light uniformly in a

den301095 [7]

Answer:

a. 0.332 W/m² b. 11.2 V/m c. 15.8 V/m

Explanation:

(a) the average intensity of the light,

Intensity, I = P/A where P = average power = 150.0 W and A = area through which the power emits = 4πr² where r = distance from bulb = 6 m.

So, I = P/A = P/4πr²

Substituting the values of the variables into the equation, we have

I = P/4πr²

I = 150.0 W/4π(6 m)²

I = 150.0 W/4π(36 m²)

I = 150.0 W/452.39 m²

I = 0.332 W/m²

(b) the rms value of the electric field,

Since Intensity, I = E²/cμ₀ where E = rms value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = permeability of free space = 4π × 10⁻⁷ H/m.

Making E subject of the formula, we have

E² = Icμ₀

E = √(Icμ₀)

Since I = 0.332 W/m², substituting the other terms into the equation, we have

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(12.5 × 10)

E = √125 V/m

E = 11.18 V/m

E ≅ 11.2 V/m

(c) the peak value of the electric field.

The peak value of electric field, E' is gotten from E = E'/√2 where E = rms value of electric field.

So, E' = √2E