A. The Sun is the only star in the Solar System. This is the only star we know of.
I hope this helps!!
Answer:
y = 1.19 m and λ = 8.6036 10⁻⁷ m
Explanation:
This is a slit interference problem, the expression for destructive interference is
d sin θ = m λ
indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m
λ = d sin θ / m
let's calculate
λ = 4.50 10⁻⁶ sin 35 /3
λ = 8.6036 10⁻⁷ m
for the separation distance from the central stripe, we use trigonometry
tan θ= y / L
y = L tan θ
the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits
L = 2 -0.30
L = 1.70 m
let's calculate
y = 1.70 tan 35
y = 1.19 m
They do because if one thing is moving and the is there both moveing in once
Answer:
Explanation:
a. Landing height is
H=1.3m
Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft
u=1.3m/s
Velocity of lander at impact, i.e final velocity is needed
v=?
The acceleration due to gravity is 0.4 times that of the one on earth,
Then, g on earth is approximately 9.81m/s²
Then, g on Mars is
g=0.4×9.81=3.924m/s²
Then using equation of motion for a free fall body
v²=u²+2gH
v²=1.3²+2×3.924×1.3
v²=1.69+10.2024
v²=11.8924
v=√11.8924
v=3.45m/s
The impact velocity of the spacecraft is 3.45m/s
b. For a lunar module, the safe velocity landing is 3m/s
v=3m/s.
Given that the initial velocity is 1.2m/s²
We already know acceleration due to gravity on Mars is g=3.924m/s²
The we need to know the maximum height to have a safe velocity of 3m/s
Then using equation of motion
v²=u²+2gH
3²=1.2²+2×3.924H
9=1.44+7.848H
9-1.44=7.848H
7.56=7.848H
H=7.56/7.848
H=0.963m
The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m
