1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km
<h3>
What is Speed ?</h3>
Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s
In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.
What are the parameters to consider ?
The parameters are;
- The distance S = 3.85 ×
km
- The Speed of Light C = 3 ×
m/s
Speed = distance S ÷ Time t
Convert kilometer to meter by multiplying it by 1000
C = S/t
3 × = 3.85 × / t
Make t the subject of formula
t = 3.85 × / 3 ×
t = 1.2833
t = 1.3 s
Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km
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Answer:
I t hi ink the answer is a
Explanation:
I hoped this helpes
Answer:
The speed of the plank relative to the ice is:
Explanation:
Here we can use momentum conservation. Do not forget it is relative to the ice.
(1)
Where:
- m(g) is the mass of the girl
- m(p) is the mass of the plank
- v(g) is the speed of the girl
- v(p) is the speed of the plank
Now, as we have relative velocities, we have:
(2)
v(g/b) is the speed of the girl relative to the plank
Solving the system of equations (1) and (2)
I hope it helps you!
Answer:
a) μ = 0.1957
, b) ΔK = 158.8 J
, c) K = 0.683 J
Explanation:
We must solve this problem in parts, one for the collision and the other with the conservation of energy
Let's find the speed of the wood block after the crash
Initial moment. Before the crash
p₀ = m v₁₀ + M v₂₀
Final moment. Right after the crash
pf = m 
+ M v_{2f}
The system is made up of the block and the bullet, so the moment is preserved
p₀ = pf
m v₁₀ = m v_{1f} + M v_{2f}
v_{2f} = m (v₁₀ - v_{1f}) / M
v_{2f} = 4.5 10-3 (400 - 190) /0.65
v_{2f} = 1.45 m / s
Now we can use the energy work theorem for the wood block
Starting point
Em₀ = K = ½ m v2f2
Final point
Emf = 0
W = ΔEm
- fr x = 0 - ½ m v₂₂2f2
The friction force is
fr = μN
With Newton's second law
N- W = 0
N = Mg
We substitute
-μ Mg x = - ½ M v2f2
μ = ½ v2f2 / gx
Let's calculate
μ = ½ 1.41 2 / 9.8 0.72
μ = 0.1957
b) let's look for the initial and final kinetic energy
K₀ = 1/2 m v₁²
K₀ = ½ 4.50 10⁻³ 400²
K₀ = 2.40 10² J
Kf = ½ 4.50 10⁻³ 190²
Kf = 8.12 10¹ J
Energy reduction is
K₀ - Kf = 2.40 10²- 8.12 10¹
ΔK = 158.8 J
c) kinetic energy
K = ½ M v²
K = ½ 0.650 1.45²
K = 0.683 J
