Okay, so they want to basically Increase their grip, and they are taking advantage of the force of friction
Answer:
0.68 m
Explanation:
We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s
V=frequency*wavelength
Then wavelength is given by 350/500=0.68 m
Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
Explanation:
It is given that,
Mass of the soccer ball, m = 0.425 kg
Speed of the ball, u = 15 m/s
Angle with horizontal, 
Time for which the player's foot is in contact with it, 
Part A,
The x component of the soccer ball's change in momentum is given by :



The y component of the soccer ball's change in momentum is given by :



Hence, this is the required solution.