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lorasvet [3.4K]
3 years ago
10

An elevator filled with passengers has a mass of 1603 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 f

or 2.50 s. Calculate the tension in the cable (in N) supporting the elevator.
Physics
1 answer:
Galina-37 [17]3 years ago
8 0

Answer:

T = 17649.03 N = 17.65 KN

Explanation:

The tension in the cable must be equal to the apparent weight of the passenger. For upward acceleration:

T = W_A = m(g+a)\\

where,

T = Tension in cable = ?

W_A = Apparent weight

m = mass = 1603 kg

g = acceleration due to gravity = 9.81 m/s²

a = acceleration of elevator = 1.2 m/s²

Therefore,

T = (1603\ kg)(9.81\ m/s^2+1.2\ m/s^2)\\\\

<u>T = 17649.03 N = 17.65 KN</u>

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Mathphys Help help help
Keith_Richards [23]

Answer:

1030 mph

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An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower. The electric field amplitude of the sign
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Answer:

Explanation:

Given that,

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Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

KE = \frac{1}{2}mv^2

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

U_s = \frac{1}{2}kx^2

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E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2

We can simplify and solve for 'x'.

mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}

Plug in the givens and solve.

x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}

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