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dangina [55]
4 years ago
11

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3460-kg space tug and a 6430

-kg asteroid. They pull on the asteroid with a force of 366 N. Initially the tug and the asteroid are at rest, 493 m apart. How much time does it take for the ship and the asteroid to meet?
Physics
1 answer:
alexandr1967 [171]4 years ago
8 0

Answer:

77.8s

Explanation:

Let d distance between the asteroid and space tug

So;d=Xtug+Xspace

Xtug=VtT+0.5atT^2

Xspace=VsT+0.5asT^2

Since Vt=Vs=0 initial velocity

Then

d=0.5(atT^2+asT^2)

T^2( at+as)=2d

T=√(2d/at+as)

But force F = mass M*acceleration a

Hence at=Ft/mt ,as=Fs/ms

But note Ft=F=Fs since the Same force acts on it

Hence T=√( 2d/F(1/mt+1/Ms))

T=√(2*493/366(1/3460+1/6430)

T=√(986/0.1627)=√(6060.195)=77.8s

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When this circuit is closed, which way do the electrons flow?
gizmo_the_mogwai [7]

Answer:

from the positive end of the battery through the capacitor through the resistors to the negative end...

Current flows from higher potential (+) to lower potential (-)..

3 0
3 years ago
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A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
Bumek [7]

Answer:

a) v1 = 5.52m/s

b) v2 = -1.52m/s

c) v3 = 4.62m/s

d) vt = 3.85m/s

Explanation:

The velocity of the football wide receiver is his displacement per unit time.

Velocity v = (displacement d)/time t

v = d/t .....1

For each of the cases, equation 1 would be used to calculate the velocity.

a) v1 = d1/t1

d1= 16m

t1 = 2.9s

v1 = 16m/2.9s

v1 = 5.52m/s

b) v2 = d2/t2

d2 = -2.5m

t2 = 1.65s

v2 = -2.5/1.65

v2 = -1.52m/s

c) v3 = d3/t3

d3 = 24m

t3 = 5.2s

v3 = 24/5.2

v3 = 4.62m/s

d) vt = dt/tt

dt = 16m - 2.5m + 24m = 37.5m

tt = 2.9 + 1.65 + 5.2 = 9.75s

vt = 37.5/9.75

vt = 3.85m/s

5 0
4 years ago
Light moves at a speed of around 1 million miles per hour<br>O<br>True<br>False​
AURORKA [14]

Answer:

False

Explanation:

In miles per hour, light speed is about 670,616,629 mph

5 0
3 years ago
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It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T
tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\

<u>ω = 0.05 rad/s</u>

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3 years ago
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