If a 5-solar-mass black hole has a radius to its event horizon of 15 km, the radius of the event horizon of a 50-solar-mass blac
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(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.
(b) The force the ground exerts on the front set of wheels is 0.239 MN.
<h3>Center mass of the airplane</h3>The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.
<h3>Some assumptions</h3>- The wheels under the wind do not pass through the center line.
- The position of the front wheel is constant and it is zero mark (origin).
- The rear wheels are at 21.7 m mark
Position of the center mass of the plane is calculated as follows;
Let the position of the center mass, Xcm = y
the center mass is 3 m in front of rear wheels, that is
21.7 - y = 3
y = 21.7 - 3
y = 18.7 m
Xcm = 18.7 m
<h3>Mass of the plane at the position of the rear wheels</h3>Let the mass of the plane at front wheels = M1
Let the mass of the plane at rear wheels = M2
There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;
M1 + M2 = 177,000
M1 = 177,000 - M2
M1 = 177,000 - 152,529.95
M1 = 24,470.05 kg
<h3>Force exerted by the ground on the front wheel</h3>W = mg
W = 24,470.05 x 9.8
W = 239,806.5 N = 0.239 MN
Learn more about center mass here: brainly.com/question/13499822
Answer:
P = 0.25 W
Explanation:
Given that,
The emf of the battry, E = 2 V
The resistance of a bulb, R = 16 ohms
We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :
So, 0.25 W power is delivered to the bulb.