Answer:
maybe its heat sorry if it's wrong
because if friction is not in the problem so we are making heat or thermal energy
Answer:
The wavelength of the infrared wave is <u>0.0001 m</u>.
Explanation:
Given:
Frequency of an infrared wave is, 
We know that, infrared waves are electromagnetic waves. All electromagnetic waves travel with the same speed and their magnitude is equal to the speed of light in air.
So, speed of infrared waves coming from the Sun travels with the speed of light and thus its magnitude is given as:
Where, 'v' is the speed of infrared waves and 'c' is the speed of light.
Now, we have a formula for the speed of any wave and is given as:
Where, 
Now, rewriting the above formula in terms of wavelength, 
, we get:
Now, plug in 
for 'v', 
for 'f' and solve for . This gives,
Therefore, the wavelength of the infrared wave is 0.0001 m.
Answer:
λ = 5940 Angstroms
Explanation:
This is an exercise of the relativistic Doppler effect
f’= f √((1- v / c) / (1 + v / c))
Where the speed in between the strr and the observer is positive if they move away
Let's use the relationship
c = λ f
f = c /λ
We replace
c /λ’ = c /λ √ ((1- v / c) / (1 + v / c))
λ = λ’ √ ((1- v / c) / (1 + v / c))
Let's calculate
v = 0.01 c
v = 0.01 3 10⁸
v= 3 10⁶ m / s
λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]
λ = 6000 √ [0.99 / 1.01]
λ = 5940 Angstroms
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
