Answer:
(a) Friction force = 50 N
(b) Work done by friction = 300 j
(c) Net work done = 0 j
Explanation:
We have given that the box is pulled by 6 meter so d = 6 m
Force applied on the box F = 60 N
We have have given that velocity is constant so acceleration will be zero
So to applied force will be utilized in balancing the friction force
So friction force 
Work done by friction force 
Work done by applied force 
So net work done = 300-300 = 0 j
Answer:
It is equal to the overall momentum before collision, so far no external object is involved.
Explanation:
Momentum is always conserved during collision as a rule. This is equal to the product of the mass and velocity. Thank you.
Answer:
44.3 m/s
Explanation:
a) Draw a free body diagram of the mass M. There are three forces:
Weight force mg pulling down,
Normal force N pushing perpendicular to the ramp,
and tension force T pulling parallel up the ramp.
Sum of forces in the parallel direction:
∑F = ma
T − Mg sin 30° = 0
T = Mg sin 30°
T = Mg / 2
Draw a free body diagram of the hanging mass m. There are two forces:
Weight force mg pulling down,
and tension force T pulling up.
Sum of forces in the vertical direction:
∑F = ma
T − mg = 0
T = mg
Substitute:
mg = Mg / 2
m = M / 2
M = 2m
b) Velocity of a standing wave in a string is:
v = √(T / μ)
T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N. Therefore:
v = √(49 N / 0.025 kg/m)
v = 44.3 m/s
Answer:
1.28
Explanation:
If you want to find the m/s you would divide distance by time, so
45 divided by 35 would equal 1.28571429 and so on.
you can just write the three first numbers.
Multiplied by; speed = distance x time
