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3 years ago

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An 850 kg elevator is accelerating upwards at 3.0 m/s2. Calculate the tension in the cable.

1 answer:

sveta [45]3 years ago

7 0

Answer:

10880N

Explanation:

We are given that

Mass of elevator, m=850 kg

Acceleration, $a=3m/s^2$

$g=9.8m/s^2$

We have to find the tension in the cable.

We know that

When elevator is accelerating upward then tension in the cable

$T=m(a+g)$

Using the formula

$T=850(3+9.8)$

$T=10880N$

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denis23 [38]

Answer:

$1.07004\times 10^{-7}\ m$

Explanation:

$n_s$ = Refractive index of bubble = 1.33

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The wavelength of light is given by

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Wavelength is also given by

$\lambda=\dfrac{c}{f}$

m = 1 for minimum thickness

$\dfrac{c}{f}=\dfrac{2n_st}{m-\dfrac{1}{2}}\\\Rightarrow t=\dfrac{m-\dfrac{1}{2}c}{2n_sf}\\\Rightarrow t=\dfrac{(1-\dfrac{1}{2})\times 3\times 10^8}{2\times 1.33\times 5.27\times 10^{14}}\\\Rightarrow t=1.07004\times 10^{-7}\ m$

The minimum thickness is $1.07004\times 10^{-7}\ m$

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3 years ago

A chunk of metal has a volume of

Alla [95]

Answer:

7213.7kg/m³

Explanation:

Given parameters:

Volume of the metal chunk = 0.131m³

Mass of the metal chunk = 945kg

Unknown:

Density of the metal chunk = ?

Solution :

To solve this problem, density is the mass per unit volume. It is mathematically expressed as;

Density = $\frac{mass}{volume}$

So;

Density = $\frac{945}{0.131}$ = 7213.7kg/m³

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saad has mass 80kg when resting on the ground at the equator what will be the centripetal acceleration on saad if the radius of

oksano4ka [1.4K]

I did try to solve. I hope it is correct, below is the solution:

<span>put everything in s.i units
then the answer what u wrote is acceleration to get is divide by mass(80) G=6.011*10^-11
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3 years ago

8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a

Vitek1552 [10]

Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61 N

work = 62.61 * 2 = 125.22 j

Net Force downplane = force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

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3 years ago

On the earth, an astronaut throws a ball straight upward; it stays in the air for a total time of 2.5 s before reaching the grou

lesya [120]

<h2>Answer: 15 s</h2>

Explanation:

This is a situation related to vertical motion with constant acceleration, where the equation that will be usefull is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the final height of the ball (when it reaches the ground)

$y_{o}=0$ is the initial height of the ball (is also zero because is thrown from ground)

$V_{o}$ is the initial velocity of the ball

$t=2.5 s$ is the time the ball is in air (on Earth)

$g_{E}=9.8 m/s^{2}$ is the acceleration due to gravity on Earth

$g_{M}=1.62 m/s^{2}$ is the acceleration due to gravity on the Moon

Having this clear, let's solve (1) for the Earth:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (2)

$0=t(V_{o}-\frac{1}{2}gt^{2})$ (3)

$V_{o}=\frac{1}{2}gt^{2})$ (4)

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2.5 s)^{2})$ (5)

$V_{o}=12.25 m/s$ (6) This is the initial velocity

Using this same velocity and equation (4) for the Moon:

$12.25 m/s=\frac{1}{2}(1.62 m/s^{2})t^{2})$ (7)

Finally finding $t$ :

$t=15.12 s \approx 15 s$

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3 years ago