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BabaBlast [244]
3 years ago
14

two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f

orward on the rope by moving hand over hand and gradually reeling in more of the rope. In doing so, he exerts a force of 35N [backwards] on the rope. This causes him to accelerate toward the other person. Assuming that the friction acting on the skaters is negligible, how long will it take for them to meet? Explain your reasoning.
Physics
1 answer:
Zina [86]3 years ago
5 0

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

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Diferencie energia cinética de energia potencial gravitacional.
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Answer:

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4 0
3 years ago
Ashkon throws a basketball across the court to his teammate. The ball has 57 J of potential energy and 61 J of kinetic energy. W
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The total mechanical energy of the ball is the sum of its potential energy U and its kinetic energy K, therefore:
E=U+K=57 J+61 J=118 J
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8 0
3 years ago
Read 2 more answers
A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o
maxonik [38]
For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
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4 0
3 years ago
HELP PLEASE I need to finish this asap
ELEN [110]

Answer:

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Explanation:

pls vote brainliest

6 0
2 years ago
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A stone is thrown, run a velocity of 15m/s is projected of an elevation of 30° to the horizontal calculate the time rate of flig
frozen [14]

Answer:

1.53 seconds

Explanation:

Applying,

T = 2usin∅/g................ Equation 1

Where, T = time of flight, u = initial velocity, ∅ = angle of projectile to the horizontal, g = acceleration due to gravity

From the question,

Given: u = 15 m/s, ∅ = 30°

Constant: g = 9.8 m/s²

Substitute these values in equation 1

T = 2(15)(sin30°)/9.8

T = 15/9.8

T = 1.53 seconds

Hence the time rate of flight is 1.53 seconds

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2 years ago
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