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AleksAgata [21]
3 years ago
12

A 0.145 kg baseball is thrown with a velocity of 40 m/s. What is the baseball's momentum?

Physics
1 answer:
Rom4ik [11]3 years ago
3 0
The momentum, p, is the product of the mass and the velocity:
p = m*v = 0.145 kg * 40 m/s = 5.8 kg*m/s
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Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

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a = 733.04/10

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B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

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Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

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