Answer:
16200 km/s
270 km/min
4.5 km/h
Explanation:
Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)
Simply plug in our known variables and solve:
a = (45.0 - 0)/10
a = 45.0/10
a = 4.5 km/h
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Answer:
The angle of recoil electron with respect to incident beam of photon is 22.90°.
Explanation:
Compton Scattering is the process of scattering of X-rays by a charge particle like electron.
The angle of the recoiling electron with respect to the incident beam is determine by the relation :
....(1)
Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant,
is mass of electron, c is speed of light and f is the frequency of the x-ray photon.
We know that, f = c/λ ......(2)
Here λ is wavelength of x-ray photon.
Rearrange equation (1) with the help of equation (1) in terms of λ .

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for
, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ and 134° for θ in the above equation.


= 22.90°
Answer:
1. Acceleration
Explanation:
Newtons Second law gives the measure of acceleration