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USPshnik [31]
3 years ago
5

An astronaut whose mass is 80. kg and is initially at rest carries an empty oxygen tank with a mass of 10. kg. He throws the tan

k away from himself with a speed of 2.0 m/s. With what velocity does he start to move off into spa
Physics
1 answer:
sergeinik [125]3 years ago
6 0

Answer:

0.25 m/s

Explanation:

From the law of conservation of momentum

Mu+ mu = Mv' + m v

M= mass of the astronaut = 80 kg

m= mass of the oxygen tank= 10 Kg

v= speedof the tank 2 m/s

u= initial velocity of the system= 0

If we substitute the values, we have

( 80× 0 )+(10×0)= [(80 x v )+ (10 x 2)]

0= 80v + 20

-20=80v

v= -0.25 m/s ( we have a negative value because the astronaut and the motion of the cylinder are in opposite direction)

Hence the velocity the astronaut start to move off into space is 0.25 m/s

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Albert Einstein, in his theory of special relativity, determined that the laws of physics are the same for all non-accelerating observers, and he showed that the speed of light within a vacuum is the same no matter the speed at which an observer travels

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4 years ago
In a nuclear power plant, uranium atoms ______
motikmotik

Answer:

All nuclear power plants use nuclear fission, and most nuclear power plants use uranium atoms. During nuclear fission, a neutron collides with a uranium atom and splits it, releasing a large amount of energy in the form of heat and radiation. More neutrons are also released when a uranium atom splits.

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3 years ago
Example of natural magnet​
jeyben [28]

Answer: lodestone

An example of a natural magnet is the lodestone, also called magnetite. Other examples are pyrrhotite, ferrite, and columbite.

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Explanation:

4 0
3 years ago
In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s
Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle =  35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ   with x axis    .............1

m1v3sin35 = m2v4sinθ                   with y axis  .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ  = 3.38                                            .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ  

v4 sinθ = 1.95                                              ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0
3 years ago
(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you ca
Colt1911 [192]

a) 52.5 m

b) 16.0 m/s

Explanation:

a)

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

v^2-u^2=2as

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

a=\frac{-\mu mg}{m}=-\mu g

where:

\mu=0.80 is the coefficient of friction

g=9.8 m/s^2 is the acceleration due to gravity

Substituting and solving for s, we find:

s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m

b)

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

\mu=0.25

Here we want the stopping distance of the car to remain the same as part a), so

s=52.5 m

We can use again the same suvat equation:

v^2-u^2=2as

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s

7 0
3 years ago
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