Answer:
2.26 s
Explanation:
Let's take down to be positive.
Given (in the y direction):
Δy = 25 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
25 m = (0 m/s) t + ½ (9.8 m/s²) t²
25 = 4.9t²
t = 2.26 s
If the ball instead had an initial horizontal velocity of 5 m/s, its initial vertical velocity is still 0 m/s. So the time to fall is still 2.26 s.
Answer:8 m/s
Explanation:
Given
kinetic Energy of 
initially 
is at rest and let say 
is moving with velocity u
kinetic Energy of 
In Completely inelastic collision both mass stick together and move with common velocity
Suppose v is the common velocity
therefore Final velocity with which both blocks moves is 1 m/s
Answer:
1.25 m/s
Explanation:
Given,
Mass of first ball=0.3 kg
Its speed before collision=2.5 m/s
Its speed after collision=2 m/s
Mass of second ball=0.6 kg
Momentum of 1st ball=mass of the ball*velocity
=0.3kg*2.5m/s
=0.75 kg m/s
Momentum of 2nd ball=mass of the ball*velocity
=0.6 kg*velocity of 2nd ball
Since the first ball undergoes head on collision with the second ball,
momentum of first ball=momentum of second ball
0.75 kg m/s=0.6 kg*velocity of 2nd ball
Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg
=1.25 m/s
Your answer is 3 ( 1 calcium atom and 2 bromine atoms)
It appears to be a <span>spiral shape. </span>
