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mixas84 [53]
3 years ago
5

A person weighing 0.6 kN rides in an elevator that has a downward acceleration of 1.3 m/s 2 . The acceleration of gravity is 9.8

m/s 2 . What is the magnitude of the force of the elevator floor on the person? Answer in units of kN.
Physics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer:

520.41 N.

Explanation:

From newton's second law of motion,

The force of the elevator floor on a person is given as,

R = m(g-a)

R = W-ma................................... Equation 1

Where R = Force of the elevator floor, m = mass of the person, a = acceleration of the elevator, W = weight of the person.

But,

W = mg,

where g = acceleration due to gravity.

m = W/g.................... Equation 2

Given: W = 0.6 kN, = 600 N, g = 9.8 m/s²

m = 600/9.8

m = 61.22 kg.

Given: W = 0.6 kN = 600 N, m = 61.22 kg, a = 1.3 m/s²

Substitute into equation 1

R = 600-61.22(1.3)

R = 600-79.586

R = 520.41 N.

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4 0
3 years ago
A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist
galina1969 [7]

Answer:

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Explanation:

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Force is given by

F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N

When we multiply force and radius we get torque

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\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm

Torque on forefinger

\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm

The total torque is given by

\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm

The most torque that exerted on the knob is 0.04225 Nm

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