A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 555 N. As the drawing sh
ows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tension in the rope to the left of the mountain climber.
1 answer:
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The tension in the rope on the left of the mountain climber is
Explanation:
From the question we are told that
The weight of the mountain climber is m = 555 N
Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as
Here are the tension of the rope on the left and on the right hand side
So
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Generally from the diagram , the total amount of force acting on the rope along the horizontal axis at equilibrium is mathematically represented as
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Answer:
λ = 5.85 x 10⁻⁷ m = 585 nm
f = 5.13 x 10¹⁴ Hz
Explanation:
We will use Young's Double Slit Experiment's Formula here:
where,
λ = wavelength = ?
Y = Fringe Spacing = 6.5 cm = 0.065 m
d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m
L = screen distance = 5 m
Therefore,
<u>λ = 5.85 x 10⁻⁷ m = 585 nm</u>
Now, the frequency can be given as:
where,
f = frequency = ?
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>f = 5.13 x 10¹⁴ Hz</u>