Displacement = 0, assuming that he runs back to original position
Average velocity is displacement/ time, since displacement =0, velocity is also 0
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:
For communication to be communicated between you and your cousin who is in Houston,Texas, when a call is made by you, a request is made to the specific phone, and the telephone tower will get the request from the mobile phone. Then a signal is sent via a transmitter underground, by this the satellite communicates with the local receiver in Houston, that is linked to the local tower over there, the tower would request for your cousin's number and connects the two of you, once a link is set.
Explanation:
From the example stated, what is required for such for a far distance, is a communication satellite link.
When a call is made by you, the a connection request is sent to the specified phone.The telephone tower receives the request from The mobile phone. The local tower(Birmingham,Al) is linked to a ground transmitter by the means of a Fiber optical cable.
A signal is sent to satellite via the ground transmitter.The satellite then set's off the local receiver in (Houston,Texas) which on it's end is connected to the local tower there. This tower then ask for your cousin's mobile for a call that will be incoming, a link is set, once he/she receives the call, from there a conversation can be done.
To solve this problem, we use the equation:
<span>d = (v^2 - v0^2) /
2a</span>
where,
d = distance of collapse
v0 = initial velocity = 101 km / h = 28.06 m / s
v = final velocity = 0
a = acceleration = - 300 m / s^2
d = (-28.06 m / s)^2 / (2 * - 300 m / s^2)
<span>d = 1.31 m</span>
