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lubasha [3.4K]
2 years ago
13

an elevator suspended by a vertical cable is moving downward but slowing down. the tension in the cable must be:

Physics
1 answer:
vladimir1956 [14]2 years ago
6 0

Answer:

F = M a is the vector equation involved

F = T - M g      are the forces acting on the elevator   (scalar equation)

T - M g = M a

T = M (a + g)    remember this a scalar

If a is slowing down then it must have a positive acceleration upwards

Therefor the tension in the cable must be greater than zero

When the tension increases to M g, a has increased to zero

For a to be zero, no acceleration, T = M g

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Find the intensity of the electromagnetic wave described in each case.

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You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is
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The initial potential energy of the wagon containing gold boxes will enable

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The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

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Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

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The answer is Corona 
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3 years ago
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A water wave has a speed of 23.0 meters/second. If the wave’s frequency is 0.0680 hertz, what is the wavelength?
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Letter B
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3 years ago
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Answer:

19.6N

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Given parameters:

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Unknown:

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6 0
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