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timofeeve [1]
3 years ago
11

To a stationary, a bus moves south with a speed of 15 m/s. a man inside walks toward the front of the bus with a speed of 0.2 m/

s relative to the bus. What is the velocity of the man according to a stationary observer?
Physics
2 answers:
marta [7]3 years ago
8 0
It is 15.2 m/s South.
Alenkasestr [34]3 years ago
3 0

Answer:

Velocity of man with respect to stationary observer is 15.2 m/s

Explanation:

It is given that, to a stationary, a bus moves south with a speed of 15 m/s. A man inside walks toward the front of the bus with a speed of 0.2 m/s relative to the bus. We have to find the velocity of man according to stationary observer.

The velocity of bus and man are in same direction. So, the velocity of man with respect to stationary observer will be :

v_{m,o}=v_b+v_m

v_{m,o}=15\ m/s+0.2\ m/s

v_{m,o}=15.2\ m/s

Hence, this is the required solution.                 

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A ball with a momentum of 16 kg.M/s strikes a ball at rest. What is the total momentum of both the balls after the collision.
melomori [17]

Answer:

Total momentum = 16 Kgm/s

Explanation:

Let the momentum of the two balls be A and B respectively.

Momentum A = 16 kgm/s

Momentum B = 0 kgm/s (since the ball is at rest).

Total momentum = A + B

Total momentum = 16 + 0

Total momentum = 16 Kgm/s

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

8 0
2 years ago
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Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

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x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

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