"Wheel & Axle" <span>can be described as a shaft that is attached to the center of a wheel
Hope this helps!</span>
F = ma
F = force
m = mass
a = acceleration
4.2 m/s = a
70 kg = m
F = ?
F = 4.2*70
F = 294 Newtons
Answer would then be : Force = 294 N
Answer:
The frequency of the photon decreases upon scattering
Explanation:
Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.
Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.
Therefore, since the wavelength increases, we have from
λf = λ'f' = c
f = c/λ
Where:
f and f' = The frequency of the motion of the photon before and after the scattering
c = Speed of light (constant)
We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;
f = c/λ
As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.
Answer:
2.136 N-m
Explanation:
From the question,
Applying,
Toque = FL...................... Equation 1
Where F = Force applied to the the ball, L = Length of the forearm
From newton's second law of motion,
F = m(v-u)/t................. Equation 2
Where m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time.
Given: m = 0.140 kg, u = 0 m/s (from rest), v = 24 m/s, t = 0.425 s
Substitute into equation 2
F = 0.140(24-0)/0.425
F = 7.91 N.
Also Given: L = 0.270 m
Substitute into equation 1
Torque = 7.91×0.270
Torque = 2.136 N-m
Answer:
the magnitude of the angular magnification of the telescope. is 4
Explanation:
Calculate the magnitude of the angular magnification of the telescope.
Given that,
distance = 25cm
focal length from the objective f₀ = 20cm
focal length from eye piece f₁ = 5cm
The angular magnification of the telescope is
Magnification = 20 / 5
magnification = 4
Hence, the magnitude of the angular magnification of the telescope. is 4