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kondaur [170]
4 years ago
5

The elevator accelerates upward (in the positive direction) from rest at a rate of 1.95 m/s2 for 2.15 s. Calculate the tension i

n the cable supporting the elevator in newtons.
Physics
1 answer:
sammy [17]4 years ago
3 0

The mass is missing. The mass of the elevator is 1650 kg.

Answer:

The tension in the cable is 19387.5 N.

Explanation:

Given:

Initial velocity of the elevator (u) = 0 m/s

Acceleration in the upward direction (a) = 1.95 m/s²

Time taken by the elevator (t) = 2.15 s

Mass of the elevator and persons (m) = 1650 kg

Let the tension in the cable wire be 'T' Newtons.

Now, there are 2 forces acting in the vertical direction. One is tension in the upward direction and the other the weight of the elevator in the downward direction.

As the elevator is accelerating upward, the net force acts in the upward direction.

So, net force on the elevator is given as:

F_{net}=T-mg

Now, from Newton's second law, net force equals mass times acceleration.

F_{net}=ma\\\\T-mg=ma\\\\T=m(g+a)

Plug in the given values and solve for 'T'. This gives,

T=1650\ kg(9.8+1.95)\ m/s^2\\\\T=1650\times11.75\ N\\\\T=19387.5\ N

Therefore, the tension in the cable is 19387.5 N.

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The storage coefficient of a confined aquifer is 6.8x10-4 determined by a pumping test. The thickness of the aquifer is 50 m and
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Answer

given,

storage coefficient, S = 6.8 x 10⁻⁴

thickness of aquifer, t = 50 m

porosity of the aquifer, n = 25 % = 0.25

Density of the water, γ = 9810 N/m³

Compressibilty  of water,β = 4.673 x 10⁻¹⁰ m²/N

We know,

   S = γ t(nβ + α)

where, α is the compressibility of the aquifer

   6.8 x 10⁻⁴  =9810 x 50 x (0.25 x 4.673 x 10⁻¹⁰+ α)

     α = 1.269 x 10⁻⁹ m²/N

Expansability of water

            = n t β γ

            = 0.25 x 50 x 4.673 x 10⁻¹⁰ x 9810

            = 5.73 x 10⁻⁵

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The ball rolling along a straight and level path. The ball is rolling at a constant
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Kinetic energy=0.5 x m x v^2

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the similarity of coastlines for different continents suggests that they may once have been connected,but the fact that they wer
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A lens forms an image of an object. The object is 16.0cm from the lens. The image is 12.0cm from the lens on the same side as th
Eddi Din [679]

A) -48.0 cm

In order to find the focal length of the lens, we can use the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have:

p = 16.0 cm

q = -12.0 cm (the negative sign is due to the fact that the image is on the same side as the object, so it is a virtual image, so the sign of q is negative)

Substituting, we find f:

\frac{1}{f}=\frac{1}{16}+\frac{1}{-12}=-0.020833 cm^{-1} \rightarrow f=-48 cm

B) Diverging

We have two types of lenses:

- A converging (convex) lens is curved outwards in its center, so that the incoming rays of light parallel to the principal axis are focused into the focus of the lens, on the opposite side

- A diverging (concave) lens is curved inwards in its center, so that the incoming rays of light parallel to the principar axis are deviated away from the principal axis, and they appear to come all from the focal point of the length on the same side of the object

A converging lens is identified by a positive focal length, while the focal length in a diverging lens is negative. Here, f = -48.0 cm, so this is a diverging lens.

C) 6.38 mm

We can answer this part of the problem by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the size of the image

y is the size of the object

Here we have:

y = 8.50 mm

p = 16.0 cm

q = -12.0 cm

So we find:

y' = - \frac{q}{p}y=-\frac{(-12)}{16}(8.50)=6.38 mm

D) Erect

We can determine the orientation of the image by looking at the sign of the size of the image found in part C). In fact:

- if the image is erect, the sign of y' is positive

- if the image is inverted, the sign of y' is negative

In this situation, we see that

y' = 6.38 mm

Which is positive, so the image is erect.

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