Answer:


Explanation:
what is the smallest crater that each of these telescopes could resolve on our moon?
For moon ;
s = 3.8 × 10 ⁸ m
y = 1.22 λs/D
where;
λ = 400 nm = 400× 10 ⁻⁹
D = 2.4 m
The smallest crater for the hubble space is calculated as follows:


For Aceribo ;
y = 1.22 λs/D
where :
λ = 75 cm = 0.75 m
D = 305 m


We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is

where

Now we find the moment of inertia by integrating from

to

The moment of inertia is

(from (-a/2) to

(a/2))
Answer:
35 m
0.56 m/s west
Explanation:
A) Total distance is the length of the path taken.
30 m + 5 m = 35 m
B) Velocity is displacement over time. Displacement is the difference between the final position and the initial position.
If west is -x, and east is +x, then:
Δx = -30 m + 5 m
Δx = -25 m
v = Δx / t
v = -25 m / 45 s
v = -0.56 m/s
v = 0.56 m/s west
Answer:
p = mv
m = p/v = 125000/22 = 5682 kg
Explanation:
Direct application of the momentum equation
p = mv
where,
p: momentum
m: mass
v: object velocity
steps:
-------
1) check for units consistency ( SI or Imperial)
2) separate the variable you are looking for.
3) DONE! :DD
Answer:
The diagram represents two charges, q1 and q2, separated by a distance d. Which change would produce the greatest increase in the electrical force between the two charges? *
Explanation:
doubling charge q1, only