Answer:
2.5 m/s east
Explanation:
Let east be the positive direction for velocity.
The change in momentum of the 0.75 kg model car is ...
m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)
= (0.75 kg)(20 m/s) = 15 kg·m/s
The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.
m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s
The required relation is ...
15 kg·m/s = -2(v2 -10) kg·m/s
-7.5 = v2 -10 . . . . divide by -2
2.5 = v2 . . . . . . . add 10
The velocity of the model truck after the collision is 2.5 m/s east.
Answer: 7.436 s
Explanation:
This situation is related to vertical motion, specifically free fall and can be modelled by the following equation:
Where:
is the final height of the object (when it makes splash)
is the initial height of the object
is the initial velocity of the object (it was dropped)
is the acceleration due gravity (directed downwards)
is the time since the objecct is dropped until it makes splash
Clearing
:
Finally:
10,000 meters: 1,000 meters in every kilometer.
Hence the expression of ω in terms of m and k is

Given the expressions;

Equating both expressions we will have;

Divide both equations by 2π

Square both sides

Take the square root of both sides

Hence the expression of ω in terms of m and k is

Answer:
I do belive that it is B hrs cn I an gn