Answer:
when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.
when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.
when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.
when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.
Explanation:
Th answer is A sorry if this isn’t what your looking for
Answer:

Explanation:
given,
length of the ship = 120 m
length of model of the ship = 4 m
Speed at which the ship travels = 70 km/h
speed of model = ?
by using froude's law

for dynamic similarities




hence, the velocity of model will be 12.78 km/h
Given :
Number of operations move through a pocket calculator during a full day's operation ,
.
To Find :
How many coulombs of charge moved through it .
Solution :
We know , charge in one electron is :

So , charge on n electron is :

Therefore , -21.44 coulombs of charge is moved through it .
Hence , this is the required solution .
Answer:
<em>Total momentum is conserved</em>
Explanation:
<u>Conservation of Momentum
</u>
The momentum is a physical magnitude that measures the product of the object's velocity by its mass. The total momentum of a system is the sum of all its components' individual momentums. The two-bear system starts with a total moment of

When both bears stick together, the total mass is 20 kg, and the new momentum is

We have assumed both bears move to the right after the collision. In this situation, the total momentum is conserved