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3 years ago

9

It has been suggested that rotating cylinders about 12.5 mi long and 3.99 mi in diameter be placed in space and used as colonies

. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth

1 answer:

sertanlavr [38]3 years ago

7 0

Answer:

The correct answer to the following question will be "0.0562 rad/s".

Explanation:

$r =\frac{3.9}{2}\times 1609.34$

$=3138.213\ m$

As we know,

⇒ $\omega^2 \ r=g$

On putting the values, we get

⇒ $\omega^2\times 3138.213=9.8$

⇒ $\omega = \sqrt{\frac{9.8}{3138.213}}$

⇒ $\omega = 0.0562 \ rad /s$

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A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.

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Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$

$sin \theta=\frac{6.68 m/s}{8.33 m/s}$

Clearing $\theta$ :

$\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})$

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4 years ago

A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight

photoshop1234 [79]

Answer:

$t=1.9 sec$

Explanation:

From the question we are told that:

Height $h=28m$

Time $t=3s$

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$s=ut+\frtac{1}{2}at^2$

$28=3u-\frac{1}{2}*9.8*3^2$

$u=24.03m/s$

Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28

$v=\sqrt{24.03-2*9.8*28}$

$v=5.35m/s and -5.35m/s$

Generally the Newton's equation for time to reach initial velocity is mathematically given by

$v=u+at$

$5.35=24.03-9.8t$

$t=\frac{28.03-5.35}{9.8}$

$t=1.9 sec$

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To do the same amount of work in less time you need to

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Answer:

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Explanation:

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Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision

Sidana [21]

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

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