Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
Answer: Explanation:
If the net force on an object is doubled, its acceleration will double If the mass of an object is doubled, the acceleration will be halved .
Answer: it goes the same speed as the car
Explanation:
Answer:
93 km/h
Explanation:
Given that a bus took 8 hours to travel 639 km. For the first 5 hours, it travelled at an average speed of 72 km/h
Let the first 5 hours journey distance = F
From the formula of speed,
Speed = distance/time
Substitute speed and time
72 = F/5
F = 72 × 5 = 360 km
The remaining distance will be:
639 - 360 = 279km
The remaining time will be:
8 - 5 = 3 hours
Speed = 279/3
Speed = 93 km/h
Therefore, the average speed for the remaining time of the journey is equal to 93 km/h
A) d. 10T
When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by
where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.
This force acts as a centripetal force, keeping the particle in a circular motion - so we can write
which can be rewritten as
The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):
So, we get:
We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.
B) 
The frequency of revolution of a particle in uniform circular motion is
where
f is the frequency
T is the period
We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:
T' = 10 T
Then its frequency of revolution will be:
