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2 years ago

9

un astronauta cuya masa es 80kg permanece inmovil en el espacio exterior, al activar la unidad propulsora que lleva en la espald

a, esta expulsa 2kg de gas a una rapidez de 15m/s. determina el valor de la velocidad de retroceso del astronauta

1 answer:

beks73 [17]2 years ago

5 0

Answer:

Explanation:

80(0) = 2(15) + (80-2)v

v = - 0.38 m/s

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A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

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3 years ago

A block slides down a rough ramp with a 30-degree incline as shown.

In-s [12.5K]

Answer:

Image 4 ?

Explanation:

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3 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo

yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is = $102^{\circ}-65.3^{\circ}=36.7^{\circ}$

Speed of first plane = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, $r^2$ representing the distance between the planes, we see that:

$r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295$

r = 741.6959 m

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If yall friend me ill give lots of points like this in the future

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Answer:

ok......................

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vampirchik [111]

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