337493603.8m/s²
Explanation:
Radius of the earth = 6.38 x 10¹⁶m
time = 24hr (86400s)
Unknown:
Centripetal acceleration = ?
Solution:
The centripetal acceleration is directed inward to keep the body from falling off the surface of the earth.
centripetal acceleration = 
where v is the velocity and r is the radius
also;
v = wr
where w is the angular velocity
substituting in the equation for centripetal acceleration gives;
a = w²r
also w = 
therefore;
a = 
a = 
a = 337493603.8m/s²
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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
Explanation:
Given:
v₀ = 0 m/s
v = 49 m/s
a = 9.8 m/s²
Find: t
v = at + v₀
49 m/s = (9.8 m/s²) t + 0 m/s
t = 5 s
Answer:
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
, r is radius
Let I is the displacement current. It is given by :
Here, 
is rate of increasing potential difference
So
So, the value of displacement current is 
.
