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Eduardwww [97]
3 years ago
10

Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed

of 0.624c and determines its lifetime to be 159 ns. (a) Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory
Physics
1 answer:
ANTONII [103]3 years ago
3 0

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

we know that; c is speed of light which is equal to 3 × 10⁸ m/s

we know that

distance = vt

or s = ut

so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance =  29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

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A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h
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Answer:

Explanation:

Given

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speed of bus towards North v_1=15.2 km/h\approx 4.22\ m/s

mass of bus travelling in South direction is m_2=1.578\times 10^4 kg

speed of bus v_2=12.2 km/h\approx 3.38\ m/s

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P_1=1.6\times 10^4\times 4.22

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P_2=m_2\times v_2+n\cdot m_0\times v_2

P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38

For total momentum to be towards south

P_2-P_1 should be greater than 0

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P_2=P_1

(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4

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If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
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This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

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a) What is the angular acceleration?

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Answer:

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Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

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b)

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a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

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a = 4.9 g

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