Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Answer:
A.
Explanation:
X represents the transmitting power Modulates (amplitude or frequency am/fm), amplifies the signal, transmitting it out at whatever direction the antenna is set up for.
A. Genus and B. Species are also major levels of classification.
Answer:
Explanation:
Intensity is given by the expresion:
where:
Io = inicial intensity
r1= initial distance
r= final distance
Answer:
Explanation:
Given
Required
Determine the distance at which the lighting struck
First, we need to determine the speed at which the lighting struck because the peed of sound varies with temperature.
At about 28C, the speed of sound is 346m/s
So, we have the following:
Distance is calculated as thus:
Divide by 1000 to get distance equivalent in kilometers
---- Approximated
