A) the final velocity = 66/9 m/s.
b) The total momentum before and after collision is the same because energy is destroyed or made.
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Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
Friction
Friction also affects the movement of an object on a slope. Friction is a force that offers resistance to movement when one object is in contact with another. Imagine now that you were on the downside of the object and applying force to keep the object in the same place (not moving)
6 is between 3 and 4 or 3.5 7 is 7.5 and 8 is buggy 1 will win and buggy 1 will have to wait 31.5
Answer:
Option C. 30 m
Explanation:
From the graph given in the question above,
At t = 1 s,
The displacement of the car is 10 m
At t = 4 s
The displacement of the car is 40 m
Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:
Displacement at t = 1 s (d1) = 10 m
Displacement at t= 4 s (d2) = 40
Displacement between t = 1 and t = 4 (ΔD) =?
ΔD = d2 – d1
ΔD = 40 – 10
ΔD = 30 m.
Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.