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Elza [17]
2 years ago
8

An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally

moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42 m.
Physics
1 answer:
amid [387]2 years ago
6 0

The tension in the supporting cable when the cab originally moves downward is 18422.4 N

What is tension?

Tension is described as the pulling force by the means of a three-dimensional object.

Tension might also be described as the action-reaction pair of forces acting at each end of said elements.

Here,

m =combined mass = 1600 kg

s = Displacement of the elevator = 42 m

g = Acceleration due to gravity = 9.81 m/s²

u = Initial velocity = 12 m/s

v = Final velocity = 0

According to the equation of motion:

v^{2} - u^{2} = 2as

0 - 12^2 = 2*a*42

a = - 144 / 84

a = - 1.714 m/s^2

Now let's write the equation of the forces acting on the elevator. Taking upward as positive direction:

T-mg = ma

T = m(g-a)

T = 1600 ( 9.8-(-1.74))

T=18422.4 N

Hence,

The tension in the supporting cable when the cab, originally moving downward is 18422.4 N

Learn more about tension here:

<u>brainly.com/question/13772148</u>

#SPJ4

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34kurt

Answer:

According to the Big Bang Theory, the density and temperature of the Universe is <u>lower</u> now than in the past.

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5 0
3 years ago
A 40W lamp wastes 34 J of energy every second by heating its surroundings.
Artemon [7]

Answer:

15\%.

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is 40\; \rm W (same as 40\; \rm J \cdot s^{-1},) meaning that 40\; \rm J of energy is supplied to this lamp every second.

The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive 40\; \rm J of energy input and would outputs 6\; \rm J of useful work. The efficiency of this lamp would be:

\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}.

4 0
2 years ago
A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the
Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion  that relates the height as follow:

h=v_{0} t+\frac{gt^{2}}{2}

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

h=\frac{gt^{2}}{2}

We know the height h of the spaceship hovering, and the gravity of Venus is g=8.87\frac{m}{s^{2}}. Substituting this values in the equation h=\frac{gt^{2}}{2}:

17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth g=9.81\frac{m}{s^{2}}.

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8 0
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Answer:

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Explanation:

Given

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Area = 25km² in a century.

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