Answer:
0.0133A
Explanation:
Since we have two sections, for the Inductor region there would be a current 
. In the case of resistance 2, it will cross a current 
Defined this we proceed to obtain our equations,
For ,
For ,
The current in the entire battery is equivalent to,
Our values are,
Replacing in the current for t= 0.4m/s
Answer:
Relative to the ground, the velocity of the aircraft is 240 km/hr
Explanation:
Relative velocity is different from normal velocity;
When 2 objects are moving in opposite directions towards each other, they will appear to be faster than they actually are;
This is known as the relative velocity;
The information tells us we have the aircraft moving 320 km/hr northwards relative to the wind;
The wind is in the opposite direction at 80 km/hr;
R = relative velocity of the aircraft
v = actual velocity of the aircraft
w = velocity of the wind
R = v + w
Note: if the wind was moving in the same direction, the formula would be R = v - w
320 = v + 80
v = 320 - 80
v = 240
The velocity relative to the ground is simply the actual velocity as the ground doesn't move;
So, relative to the ground, the velocity of the aircraft is simply 240 km/hr
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: True
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