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
The final velocity of the bullet+block is 0.799 m/s
Explanation:
We can solve this problem by applying the principle of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision.
Mathematically, we can write:
where
m = 0.001 kg is the mass of the bullet
u = 800 m/s is the initial velocity of the bullet
M = 1 kg is the mass of the block
U = 0 is the initial velocity of the block (initially at rest)
v is the final combined velocity of the bullet and the block
Solving the equation for v, we find the final velocity:
Learn more about conservation of momentum:
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Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d = ×t = vcos(∅)×
=
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d = =
=
=
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.