Answer:
The mass of the cart is 5 kg
Explanation:
You divide 25 by 5 and get 5. Have a great day! :D
<em>The Equation:</em>
25/5 = 5
Information that is given:
a = -5.4m/s^2
v0 = 25 m/s
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S = ?
Calculate the S(distance car traveled) with the formula for velocity of decelerated motion:
v^2 = v0^2 - 2aS
The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
S = (625 m^2/s^2)/(10.8 m/s^2)
S = 57.87 m
Answer:
5) Displacement = +3.125 m
Displacement is in the same direction as the force vector.
6) Force = -53.89 N
Force is in an opposite direction relative to the displacement.
Explanation:
5) We are given;
Force; F = 160 N.
Workdone; W = +500 J
Now, formula for workdone is;
W = Force × displacement
Thus, displacement = Work/force
Displacement = 500/160
Displacement = +3.125 m
Thus, displacement is in the same direction as the force vector.
6) We are given;
Displacement; d = 18 m.
Workdone; W = -970 J
Like in the first answer above,
Workdone = Force × Displacement
Thus;
Force = Workdone/Displacement
Force = -970/18
Force = -53.89 N
Since force is negative and displacement is positive, it means force is in an opposite direction relative to the displacement.
Answer:
Part a)

Part b)

Explanation:
As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same
So we will have




Part b)
By equation of kinetic energy we have




Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
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Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.