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
The displacement of Itzel according to the question is 6.3 miles SW
Displacement is defined as the distance moved by a body in a specified direction
Find the diagram attached
From the diagram given, we can see that AB is the displacement
To get the length AB, we will have to use the Pythagoras theorem:
![AB^2=2^2+ 6^2\\AB^2 ^2=4+36\\AB^2=40\\AB=\sqrt{40}\\AB= 6.3 miles\\](https://tex.z-dn.net/?f=AB%5E2%3D2%5E2%2B%206%5E2%5C%5CAB%5E2%20%5E2%3D4%2B36%5C%5CAB%5E2%3D40%5C%5CAB%3D%5Csqrt%7B40%7D%5C%5CAB%3D%206.3%20miles%5C%5C)
From the diagram, we can also se that the direction of the displacement in the South West direction.
Hence the displacement of Itzel according to the question is 6.3 miles SW
Learn more here: brainly.com/question/19108075
Answer:
D 9.8 m/s^2
Explanation:
The force of gravitational gravity on earth is 9.8 m/s^2
3780 km.
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Answer
2) 1.5×10-2 m
Explanation
The potential difference is related to the electric field by:
(1)
where
is the potential difference
E is the electric field
d is the distance
We want to know the distance the detectors have to be placed in order to achieve an electric field of
![E=1 V/cm=100 V/m](https://tex.z-dn.net/?f=E%3D1%20V%2Fcm%3D100%20V%2Fm)
when connected to a battery with potential difference
![\Delta V=1.5 V](https://tex.z-dn.net/?f=%5CDelta%20V%3D1.5%20V)
Solving the equation (1) for d, we find
![d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5CDelta%20V%7D%7BE%7D%3D%5Cfrac%7B1.5%20V%7D%7B100%20V%2Fm%7D%3D0.015%20m%3D1.5%20%5Ccdot%2010%5E%7B-2%7D%20m)