Question: How fast was the arrow moving before it joined the block?
Answer:
The arrow was moving at 15.9 m/s.
Explanation:
The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:
where 
is the mass of the arrow, 
is the mass of the block, 
of the change in height of the block after the collision, and 
is the velocity of the arrow before it hit the block.
Solving for the velocity , we get:
and we put in the numerical values
,
and simplify to get:
The arrow was moving at 15.9 m/s
Answer: 3 m/s
Explanation:
We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:
The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore
So, the final total momentum will also be
And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:
from which we find the velocity of the solid ball
Answer:
50N
Explanation:
Force (N) = mass (kg) × acceleration (m/s²)
0.25kg times 200m/s² = 50N
Answer:
PE = (|accepted value – experimental value| \ accepted value) x 100%
Explanation:
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
