To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.
The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r
where:
v = velocity, m/s
r= radium, m
assuming the velocity does not change:
at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
= 81 m/s^2
at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
= 16.2 m/s^2
to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill
= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
Answer:
The voltage will quadruple
Explanation:
The power dissipated in a circuit is given by
where
V is the voltage
R is the resistance
In this problem, the voltage across the circuit is doubled:
V' = 2V
So the new power dissipated is
so, the power dissipated will quadruple.
Answer:
Explanation:
Check the attachment for solution
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
