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2 years ago

I need help with those 2.

Physics

1 answer:

Nimfa-mama [501]2 years ago

5 0

Hi there!

Question 1:

We know that Impulse = Δp = mΔv, so:

I = 0.045(70) = 3.15 Ns

I = F · t, so:

3.15/0.1 = F

F = 31.5 N

Question 2.

We can find the impulse using:

I = Ft

I = 300 · 0.04 = 12 Ns

Find the change in velocity using:

I = mΔv

12/0.13 = Δv = 92.31 m/s

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3 years ago

QUESTION 3

Alisiya [41]

The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately <u>2.937 m/s²</u>.

Reason:

The given parameters are;

The mass of the wood, m = 100 kg

The force which can move the wood, F = 588 N

Wood on wood static friction, $\mu_s$ = 0.5

Wood on wood kinetic friction, $\mu_k$ = 0.3

Solution;

The force of friction, $F_f$ , acting when the crate is moving is given as

follows;

$F_f$ = m × g × $\mu_k$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore, we have;

$F_f$ = 100 × 9.81 × 0.3 = 294.3

The force of friction, $F_f$ = 294.3 N

The force with which the crate moves, F = 588 - 294.3 = 293.7

The force with which the crate moves, F = 293.7 N

Force = Mass, m × Acceleration, a

$a = \dfrac{F}{m}$

Therefore;

$a = \dfrac{293.7 \ N}{100 \ kg} = 2.937$

The acceleration of the crate, a ≈ <u>2.937 m/s²</u>.

Learn more about friction here:

brainly.com/question/94428

8 0

2 years ago

Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par

taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ = 5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

$t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 = \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s$

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0

3 years ago