Letter D: The amount of sleep a student gets affects student achievement
Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π × 
................2
put here value
period of oscillations = 2π × 
period of oscillations = 0.6953
so period of oscillations is 0.695 second
Answer:
1.125m/s^2
Explanation:
Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically
v^2= u^2+2as
Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.
a = ?
u = 0m/s
v = 15m/s
s = 100m
Substituting the values into the formula above
v^2= u^2+2as
15^2=0^2+2×a×100
225= 0+200a
225= 200a
Divide both sides by 200
225/200 = 200a/200
a= 1.125m/s^2
Hence the acceleration of the car is 1.125m/s^2.
Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s
Pressure increases with increasing depth. h2=2hh
I’ll say c Bc it make more since to find the travel distance
