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.
Learn more about moon on:
brainly.com/question/13262798
#SPJ1
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
D the distance for trial 3 will be greater than trial 4
Answer:
a) F = 1.26 10⁵ N, b) F = 2.44 10³ N, c) F_net = 1.82 10³ N directed vertically upwards
Explanation:
For this exercise we must use the relationship between momentum and momentum
I = Δp
F t = p_f -p₀
a) It asks to find the force
as the man stops the final velocity is zero
F = 0 - p₀ / t
the speed is directed downwards which is why it is negative, therefore the result is positive
F = m v₀ / t
F = 63.5 7.89 / 3.99 10⁻³
F = 1.26 10⁵ N
b) in this case flex the knees giving a time of t = 0.205 s
F = 63.5 7.89 / 0.205
F = 2.44 10³ N
c) The net force is
F_net = Sum F
F_net = F - W
F_net = F - mg
let's calculate
F_net = 2.44 10³ - 63.5 9.8
F_net = 1.82 10³ N
since it is positive it is directed vertically upwards
Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f = 
D_f = 
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m
