Answer: angular displacement in rad = 3038.45 rad
angular displacement in rev = 483.589 rev
Explanation: mathematically
Angular velocity = angular displacement / time taken.
Angular velocity = 33.5 rad/s, time taken = 90.7s
33.5 = angular displacement /90.7
Angular displacement = 33.5 * 90.7 = 3038.45 rad
But 1 rev =2π
Hence 3038.45 rad to rev is
3038.45/2π = 483.599 rev
Answer:
a) The Energy added should be 484.438 MJ
b) The Kinetic Energy change is -484.438 MJ
c) The Potential Energy change is 968.907 MJ
Explanation:
Let 'm' be the mass of the satellite , 'M'(6×
be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×
N/m) be the universal constant of gravitation.
We know that the orbital velocity(v) for a satellite -
v=
[(R+h) is the distance of the satellite from the center of the earth ]
Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)
For initial conditions ,
h =
= 98 km = 98000 m
∴Initial Energy (
) =
m
+
Substituting v=
in the above equation and simplifying we get,
= 
Similarly for final condition,
h=
= 198km = 198000 m
∴Final Energy(
) = 
a) The energy that should be added should be the difference in the energy of initial and final states -
∴ ΔE =
- 
=
(
-
)
Substituting ,
M = 6 ×
kg
m = 1036 kg
G = 6.67 × 
R = 6400000 m
= 98000 m
= 198000 m
We get ,
ΔE = 484.438 MJ
b) Change in Kinetic Energy (ΔKE) =
m[
-
]
=
[
-
]
= -ΔE
= - 484.438 MJ
c) Change in Potential Energy (ΔPE) = GMm[
-
]
= 2ΔE
= 968.907 MJ
Speed of the car given initially
v = 18 m/s
deceleration of the car after applying brakes will be
a = 3.35 m/s^2
Reaction time of the driver = 0.200 s
Now when he see the red light distance covered by the till he start pressing the brakes


Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

here
vi = 18 m/s
vf = 0
a = - 3.35
so now we will have


So total distance after which car will stop is


So car will not stop before the intersection as it is at distance 20 m
Answer:
Technician A
Explanation:
It is seen that a tire pressure will increase or decrease 1 psi for each
change in temperature.
For Technician B vehicle pressure should not be adjusted after tire has been warmed as the warm air may increase the pressure but it will be auto adjusted as the temperature falls to normal .