Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark. 
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h,  s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484 
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45) 
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
 
        
             
        
        
        
stable equilibrium, if displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement.
unstable equilibrium, if displaced it experiences a net force or torque in the same direction as the displacement from equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly.
neutral equilibrium, is when an equilibrium is independent of displacements from its original position. 
Have a good day, hope this helps
 
        
             
        
        
        
The correct answers would be B, and d
        
                    
             
        
        
        
Answer:
The time it takes the ball to fall 3.8 meters to friend below is approximately 0.88 seconds
Explanation:
The height from which the student tosses the ball to a friend, h = 3.8 meters above the friend
The direction in which the student tosses the ball = The horizontal direction
Given that the ball is tossed in the horizontal direction, and not the vertical direction, the initial vertical component of the velocity of the ball = 0
The equation of the vertical motion of the ball can therefore, be represented by the free fall equation as follows;
h = 1/2 × g × t²
Where;
g = The acceleration due gravity of the ball = 9.81 m/s²
t = The time of motion to cover height, h
Then height is already given as h = 3.8 m
Substituting gives;
3.8 = 1/2 × 9.81 × t²
t² = 3.8/(1/2 × 9.81) ≈ 0.775 s²
∴ t = √0.775 ≈ 0.88 seconds
The time it takes the ball to fall 3.8 meters to friend below is t ≈ 0.88 seconds.
 
        
             
        
        
        
The frequency of the wave has not changed.
In fact, the frequency of a wave is given by:

where v is the wave's speed and  is the wavelength.
 is the wavelength.
Applying the formula:
- In air, the frequency of the wave is:

- underwater, the frequency of the wave is:

So, the frequency has not changed.