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serg [7]
2 years ago
8

One strategy in a snowball fight is to throw

Physics
1 answer:
blondinia [14]2 years ago
7 0

Answer:

Range formula:  R = v^2 sin (2 theta) / g

If theta = 69 deg and v = 15.1

R = 15.1^2 sin 138 / 9,8 = 15.6 m

sin 138 = .669 = sin 42

So a snowball thrown at 21 deg will travel

R = 15.1 * .669^2 / 9.8 = 15.6 m

The second snowball can be thrown at 21 deg to travel the same distance

Vx = V cos theta = 15.1 * cos 69 = 5.41     first snowball

t1 = 15.6 / 5.41 = 2.88 sec

Vx = V cos theta = 15.1 cos 21 = 14.1 m/s

t2 = 15.6 / 14.1 = 1.11 sec

Difference = t1 - t2 = 1.77 sec     time delay for second snowball

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An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes
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Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

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Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = \frac{1}{2}( 60 cm - 10 cm )

A = \frac{1}{2} × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed V_{max} = ωA

and ω = 2π/T

so maximum speed V_{max} = \frac{2\pi }{T}A

so we substitute

so maximum speed V_{max} = \frac{2\pi }{3.3} × 0.25 m

so maximum speed V_{max} = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

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