Answer:
22.5 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Time (t) = 1.5 s
Final velocity (v) = 0 m/s
Distance (s) =?
The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:
s = (u + v)t/2
s = (30 + 0)1.5 / 2
s = (30 × 1.5) / 2
s = 45 / 2
s = 22.5 m
Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.
Answer:
a) 
b) 
Explanation:
Given:
String vibrates transversely fourth dynamic, thus n = 4
mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg
Tension in the string, T = 8.39 N
Length of the string, L = 1.87 m
a) we know
where,
= wavelength
on substituting the values, we get
or
b) Speed of the wave (v) in the string is given as:
also,
equating both the formula for 'v' we get,
on substituting the values, we get
or
or
Answer:
Yes, because everything bounce off in every surface around any object.
Explanation:
The second runner must run 3.3m/s. If the leading runner is 1.5 seconds ahead and there are 30m left, the second runner would need to run slightly faster than the lead in order to finish at the same time. To calculate this I did 30/1.5 which gave me 0.05. I added this onto the speed of the lead runner to get 3.3m/s :)
A drummer would loosen the drum skin in order to make the sound of a drum give a note of lower pitch.
However, I am not entirely sure.
