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erastova [34]
3 years ago
11

A dragster takes off from rest and crosses the finish line 320 m away. If the dragster is able to accelerate at 10LaTeX: \frac{m

}{s^2}m s 2 then how long does it take to finish?
Physics
2 answers:
Ber [7]3 years ago
4 0

Answer:

t = 8 s

Explanation:

Rasek [7]3 years ago
3 0

Answer:

t = 8 s

Explanation:

In order to find the time taken by the dragster we will use equations of motion. Here, we will use second equation of motion:

s = Vi t + (1/2)at²

where,

s = distance covered = 320 m

Vi = Initial Velocity = 0 m/s (Since, dragster starts from rest)

t = time taken = ?

a = acceleration of dragster = 10 m/s²

Therefore,

320 m = (0 m/s)t + (1/2)(10 m/s²)t²

t² = (320 m)(2)/(10 m/s²)

t = √(64 s²)

<u>t = 8 s</u>

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A 1300 kg steel beam is supported by two ropes. (Figure
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Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

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where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

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R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

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R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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