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2 years ago

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A force acts on a body of mass 13 kg initially at restThe force acts for 10 seconds, and once it quits, the body covers 60 m in

the next 6 seconds . Find the force that acted on the body

2 answers:

Ghella [55]2 years ago

6 0

Answer:

uguuffuufufufufucucucuvuvhvuchcyctxxtrxtccyyvuvubjbubbu

elixir [45]2 years ago

3 0

Answer: 13 N

Explanation: You have to use F=m * g

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Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

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:

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

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

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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2 years ago

**URGENT PLEASE HELP**

nalin [4]

Answer:

This depends on what angle they are approaching each other before they collided.The two simple cases are if they are running in the same direction or opposite direction from each other. For either case, use the conservation of momentum equation to solve: M_total*V_result = M1*V1 + M2*V2

Explanation:

Here are two possible solutions.

Head-on collision: M1=78, V1=8.5, M2=72, V2=-7.5 (that's negative because he's running the other way), M_total = 78+72 = 150, so V_result = (78*8.5 - 72*7.5)/150 = 0.82 m/s. Sanity check, they weigh about the same and so most of their velocity should cancel out.

Running the same way: change the sign of V2 to positive so V_result = (78*8.5 + 72*7.5)/150 = 8.02 m/s. Sanity check, they weigh about the same and the resultant speed is between the two starting velocities.

<em>hope it helps:)</em>

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