That's false. Mechanical waves (like sound and ocean waves) do
need a medium to travel in, but electromagnetic waves (like radio
and light) don't.
Answer:
The final velocity of the thrower is 
and the final velocity of the catcher is 
.
Explanation:
Given:
The mass of the thrower, 
.
The mass of the catcher, 
.
The mass of the ball, 
.
Initial velocity of the thrower, 
Final velocity of the ball, 
Initial velocity of the catcher, 
Consider that the final velocity of the thrower is 
. From the conservation of momentum,
Consider that the final velocity of the catcher is 
. From the conservation of momentum,
Thus, the final velocity of thrower is 
and that for the catcher is 
.
The answer is B I hope this helps luv
"The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Column 2 would be on the y-axis."
typical plot is speed or velocity on the y-axis n time on the x-axis so the ans is Column 1 should be titled “Time,” and Column 2 should be titled “Velocity.”
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
