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rjkz [21]
3 years ago
8

A 1000 kg elevator is accelerated upward at a rate of 0.70 m/s2. What is the tension in the cable pulling the elevator upward wh

en it experiences this acceleration?
Physics
2 answers:
Masteriza [31]3 years ago
5 0

Answer:

Tension is also known as Force...

and Force is mass× acceleration.

so....1000×0.70=700N

AlekseyPX3 years ago
3 0

Answer:

\Huge \boxed{\mathrm{700 \ N}}

Explanation:

Tension is the force exerted.

\sf Force \ (N) = mass \ (kg) \cdot acceleration \ (m/s^2)

F=1000 \cdot 0.70

F=700

The tension in the cable is 700 N.

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A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple
aliina [53]

Answer:

T= 1 s

Explanation:

Given that

When x=  cm ,T= 1

we know that time period of spring mas system given as

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So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

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5 0
3 years ago
In aircraft design, the pressure coefficient Cp is usually measured during wind tunnel testing of an aircraft component to predi
Elza [17]

Answer:

Check the explanation

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 Kindly check the attached image below for the step by step explanation to the question above.

3 0
2 years ago
How long does it take (in minutes) for light to reach venus from the sun, a distance of 1.152 × 108 km?
7nadin3 [17]
Using the precise speed of light in a vacuum (299,792,458 \ \frac{m}{s}), and your given distance of 1.152 * 10^{8} km, we can convert and cancel units to find the answer. The distance in m, using \frac{1000 \ m}{1 \ km}, is 1.152 * 10^{11} m. Next, for the speed of light, we convert from s to min, using \frac{1 \ min}{60 \ s}, so we divide the speed of light by 60. Finally, dividing the distance between the Sun and Venus by the speed of light in km per min, we find that it is 6.405 min.

7 0
3 years ago
You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

W=-Fd

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

F=\mu mg

where

\mu is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

W=-\mu mg d (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

-\frac{1}{2}mu^2 = -\mu mg d

And solving for \mu, we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767

7 0
3 years ago
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