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Temka [501]
3 years ago
5

1) A parked car starts to move with a constant velocity of 3 m/s, after moving

Physics
1 answer:
makkiz [27]3 years ago
7 0

Answer:

velocity (v) =distance(s) ÷ time (t)

Explanation:

v=s÷t

s=v×t

s=3×3.4

s= 10.2m

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Initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity
stiks02 [169]

Answer:

<em>The final velocity is 20 m/s.</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

v_f=v_o+at

The provided data is: vo=10 m/s, a=5\ m/s^2, t=2 s. The final velocity is:

v_f=10~m/s+5\ m/s^2\cdot 2\ s

v_f=20\ m/s

The final velocity is 20 m/s.

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Read 2 more answers
A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic
Ilia_Sergeevich [38]

Answer:

\mu_k=0.27

Explanation:

According to the free body diagram, in this case, we have:

\sum F_x:-F_k=ma\\\sum F_y:N=mg

Recall that the force of friction is given by:

F_k=\mu_k N

Replacing and solving for the coefficient of kinetic friction:

-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}

Finally, we calculate \mu_k:

\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27

4 0
3 years ago
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