Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
Answer:
Tha ball- earth/floor system.
Explanation:
The force acting on the ball is the force of gravity when ignoring air resistance. At the moment the player releases the ball, until it reaches the top of its bounce, the small system for which the momentum is conserved is the ball- floor system. The balls exerts and equal and opposite force on the floor. <u>Here the ball hits the floor, because in any collision the momentum is conserved. Moment of the ball -floor system is conserved</u>. Mutual gravitation bring the ball and floor together in one system. As the ball moves downwards, the earth moves upwards, although with an acceleration on the order of 1025 times smaller than that of the ball. The two objects meet, rebound and separate.
Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.
