Answer:
Explanation:
Considering that,
The mass of the first vehicle
M1= 328kg
It is traveling in the positive x direction at a speed of
U1 = 19.1m/s
The speed of the second vehicle
U2 = 13m/s, moving in the same direction as the first vehicle..
The mass of the second vehicle
M2 = 790kg
The speed of the second vehicle post-collision
V2 = 15.1 m/s
The speed of the first vehicle following the collision
V1 =?
This represents an elastic collision,
and applying the principle of conservation of momentum
The momentum prior to the collision must equal the momentum afterwards
P(before) = P(after)
M1•U1 + M2•U2 = M1•V1 + M2•V2
328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1
16534.8 = 328•V1 + 11929
328•V1 = 16534.8—11929
328•V1 = 4605.8
V1 = 4605.8/328
V1 = 14.04 m/s
The speed of the first vehicle after the collision is 14.04 m/s
Answer: Option D: indicates rapid travel with slow oscillation.
Clarification:
ycarrier(x,t) is traveling quickly but has slow oscillations.
Answer:
300 N
Explanation:
The total force exerted on the arrow is determined by Newton's Second Law:
where
m = 0.06 kg is the arrow's mass
a = 5,000 m/s² is the arrow's acceleration
By plugging in the values into the formula, we arrive at
A straightforward way to visualize this is through the relationship in a triangle: P
M V
where p represents momentum
m refers to mass
and v stands for velocity
The formulas are:
p = m x v
m = p/v
v = p/m
Thus, all provided options are correct
