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Crazy boy [7]
3 years ago
10

arlene is to walk across a high wire strung horizontally between two buildings 10.0m apart. The sag in the rope when she is at t

he midpoint is 10.0 degrees, if her mass is 50.0 kg, what is the tension in the rope at this point?...can you help me...thanks!
Physics
2 answers:
Ipatiy [6.2K]3 years ago
4 0
We are given with a force in the middle of a rope measuring 10 m and an angle of sagging of 10 degrees from the horizontal. The tension of the rope is equal to the hypotenuse of the right triangle. 

sin 10 = 50 kg * 9.8 m/s2 / T
Tension = 2821. 80 N 
Greeley [361]3 years ago
3 0

Answer:

T = 1412.3 N

Explanation:

As we know that weight of the Arlene is counterbalanced by the tension in the string

So here it is given that string makes 10 degree angle at mid point

So we will have

T sin\theta + T sin\theta = mg

so we have

T = \frac{mg}{2sin\theta}

here we know that

m = 50 kg

\theta = 10 degree

now we have

T = \frac{50 \times 9.81}{2sin10}

T = 1412.3 N

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Elis [28]

Answer:

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Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

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Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

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\Rightarrow d=5t\;\cdots(ii)

As the jogger starts from origin, so, the distance, d, also represents the position of the jogger at the time t s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

d=5\times 5=25 m

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(b) Similarly, for t=15.0 s

d=5\times 15=75 m

So, the jogger is at a distance of 75 m from the origin.

8 0
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7 0
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A bowling ball travels at 2.0 m/s. It has 16 J of kinetic energy. what Is the mass of the bowling ball in kilograms ?
zaharov [31]
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5 0
3 years ago
A ball is thrown straight upward and rises to a maximum
Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

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(0.5v_i)^2 - v_i^2 = 2(-9.81)h

-0.75v_i^2 = -19.62 h

0.75(17.72)^2 = 19.62 h

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3 0
3 years ago
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Maksim231197 [3]

Answer:

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