The number of years that will pass before the radius of the Moon's orbit increases by 3.6 x 10^6 m will be 90000000 years.
<h3>How to compute the value?</h3>
From the information given, the orbit of the moon is increasing in radius at approximately 4.0cm/yr.
Therefore, we will convey the centimeters to meter. This will be 4cm will be:
= 4/100 = 0.04m/yr.
Time = Distance / Speed
Time = 3.6 x 10^6/0.04
Time = 90000000 years.
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Complete question:
Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the moon is increasing in radius at approximately 4.0cm/y. Assuming this rate to be constant how many years will pass before the radius of the Moon's orbit increases by 3.6 x 10^6
Answer:
tan is 15 for that triangle
Answer:
Fc=5253
N
Explanation:
Answer:
Fc=5253
N
Explanation:
sequel to the question given, this question would have taken precedence:
"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."
so we derive centripetal acceleration first
ac (centripetal acceleration) = v^2/r
make r the subject of the equation
r= v^2/ac
ac is 6.23*g which is 9.81
v is 101m/s
substituing the parameters into the equation, to get the radius
(101^2)/(6.23*9.81) = 167m
Now for part
( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.
he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.
Fc (Centripetal Force) = m*v^2/r
So (86kg* 101^2)/(167) =
Fc=5253
N
Answer:
when a custodian pushes a large crate across the gym floor, forces acting on the crate are the one applied by the custodian, frictional force acting in the opposite direction of motion acting between the surface of crate and floor and reaction force (normal force) acting in upward direction normal to the surface
Explanation:
