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Lady_Fox [76]
3 years ago
13

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle

l to the ground. The mass of the stuntman is 107 kg, and the coefficient of kinetic friction between the road and him is 0.682. Find the tension in the cable.
Physics
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

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aliina [53]

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3 years ago
A tuning fork produced four beats per second with a second, 270-Hz tuning fork. What are the two possible frequencies of the fir
EleoNora [17]

Answer:

frequency of the first tuning fork can be 274 Hz or 266 Hz

Explanation:

Given data :

we have number of beats per second = 4

thus, it can also be written as

Beat frequency of the tuning fork, v = 4 Hz

Frequency of the tuning fork, f = 270 Hz

now, the beat frequency is actually the difference of the consecutive frequency.

thus, we have

Frequency of the first tuning fork as = 270 Hz + 4 Hz = 274 Hz

or

Frequency of the first tuning fork as = 270 Hz - 4 Hz = 266 Hz

Therefore, the frequency of the first tuning fork can be 274 Hz or 266 Hz

7 0
3 years ago
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maria [59]
D is the answer............
4 0
4 years ago
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Someone help me?
kap26 [50]

Answer:

-27.3 m/s

Explanation:

Given:

y₀ = 38 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2 (-9.8 m/s²) (0 m − 38 m)

v = -27.3 m/s

Or, you can solve with energy.

PE = KE

mgh = ½ mv²

v² = 2gh

v = -27.3 m/s

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Anvisha [2.4K]
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