Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
Answer:
a=2.378 m/s^2
Explanation:
a=Δv/Δt------eq(1)
Δv=Vf-Vi=120 km/h-0 km/h=120 km/h
or Δv=33.3 m/sec
or time=t=14s
putting values in eq(1)
a=33.3/14
a=2.378 m/s^2
I think the answer is A because it’s a better explanation
Answer:
acceleration of person = 9.77 m/s²
Explanation:
given data
latitude = 40 degree
to find out
Calculate the acceleration of a person
solution
we know that here 40 degree = 0.698 rad
so
acceleration of person = g - ω²R ...............1
and 1 rotation complete in 24 hours = 360 degree
here g is 9.81
so we know Earth angular speed ω = 7.27 × 
rad/s and R is earth radius that is 6.37 × 
m
so
put here value in equation 1 we get
acceleration of person = g - ω²R
acceleration of person = 9.81 - (7.27 × )² × 6.37 ×
acceleration of person = 9.77 m/s²
Protons have a positive charge which is indicated by a + sign.
Hence, the answer is C.
