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sdas [7]
3 years ago
11

You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of reference he has a speed of 3 m/s

to the south. What is the velocity of the man from the frame of reference of a stationary observer? (2 points)
Physics
1 answer:
kupik [55]3 years ago
5 0
The answer to this question would be: 1m/s

When you are walking to the north with 2m/s velocity, the stationary object(velocity=0m/s) will look like moving south at 2m/s velocity. That happens because the relative distance between you and the object is reduced by 2m/s in both conditions. In this question, the man seems like have 3m/s velocity. The real velocity should be:
3m/s - 2m/s = 1m/s
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A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then
Dominik [7]

Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

v^{2}-u^{2}= 2as

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

v^{2}-(0)^{2}= 2(1.44)(30)

v= 9.3 m^{-1}

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

s=ut+\frac{(1)}{2}at^2

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

30=(0)t+\frac{(1)}{2}(1.44)t^2

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

T=6.45+\frac{s'}{v}

Put s= 30 m, v= 9.3 m^{-1} and s'= 100 m.

T=6.45+\frac{(100)}{9.3}

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

6 0
3 years ago
The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-
just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}

6.5\text{ Light-years}=1.9929065\text{ Parsecs}

6.5\text{ Light-years}\approx 1.99\text{ Parsecs}

Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.

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