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
Do you speak a little English cuz I can’t help you if a can’t understand you
The answer is 60 mph.
The speed (v) is distance (d) per time (t): v = d/t
Car A:
v1 = ?
t1 = 2 h
d1 = ?
___
v1 = d1/t1
d1 = v1 * t1
Car B:
v2 = ?
t2 = 1.5 h
d2 = ?
___
v2 = d2/t2
d2 = v2 * t2
<span>Two cars traveled equal distances:
d1 = d2
</span>v1 * t1 = v2 * t2
<span>Car B traveled 15 mph faster than Car A:
v2 = v1 + 15
</span>v1 * t1 = v2 * t2
v2 = v1 + 15
________
v1 * 2 = (v1 + 15) * 1.5
2v1 = 1.5v1 + 22.5
2v1 - 1.5v1 = 22.5
0.5v1 = 22.5
v1 = 22.5/0.5
v1 = 45 mph
v2 = v1 + 15
v2 = 45 + 15
v2 = 60 mph
Answer:
The distance traveled in 1 year is: 
Explanation:
Given
--- speed
--- time
Required
The distance traveled
This is calculated as:
So, we have:
This gives:
-- approximated
