Answer:
option (E) 1,000,000 J
Explanation:
Given:
Mass of the suspension cable, m = 1,000 kg
Distance, h = 100 m
Now,
from the work energy theorem
Work done by the gravity = Work done by brake
or
mgh = Work done by brake
where, g is the acceleration due to the gravity = 10 m/s²
or
Work done by brake = 1000 × 10 × 100
or
Work done by brake = 1,000,000 J
this work done is the release of heat in the brakes
Hence, the correct answer is option (E) 1,000,000 J
<h3><u>Answer</u>;</h3>
= 22°
<h3><u>Explanation</u>;</h3>
- According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
- Therefore; Sin i/Sin r = η
In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33
Thus;
Sin 30 / Sin r = 1.33
Sin r = Sin 30°/1.33
= 0.3759
r = Sin^-1 0.3759
= 22.08
<u>≈ 22°</u>
Answer: 80m
Explanation:
Distance of balloon to the ground is 3150m
Let the distance of Menin's pocket to the ground be x
Let the distance between Menin's pocket to the balloon be y
Hence, x=3150-y------1
Using the equation of motion,
V^2= U^s + 2gs--------2
U= initial speed is 0m/s
g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s
40m/s is contant since U (the coin is at rest is 0) hence V =40m/s
Slotting our values into equation 2
40^2= 0^2 + 2 * 10* (3150-y)
1600 = 0 + 63000 - 20y
1600 - 63000 = - 20y
-61400 = - 20y minus cancel out minus on both sides of the equation
61400 = 20y
Hence y = 61400/20
3070m
Hence, recall equation 1
x = 3150 - 3070
80m
I hope this solve the problem.
Answer:
v_average = 500 m / min
Explanation:
Average speed is defined
v = (x_{f} -x₀) / Δt
let's look in each section
section 1
the variation of the distance is 800 in a time of 1.4 min
v₁ = 800 / 1.4
v₁ = 571.4 m / min
section 2
distance interval 500 in a 1.6 min time interval
v₂ = 500 / 1.6
v₂ = 312.5 m / min
section 3
distance interval 1200 m in a time 2 min
v₃ = 1200/2
v₃ = 600 m / min
taking the speed of each section we can calculate the average speed
the distance traveled
Δx = 800 + 500 + 1200
Δx = 2500 m
the time spent
Δt = 1.4 + 1.6+ 2
Δt = 5 min
v_average = Δx / Δt
v_average = 2500/5
v_average = 500 m / min
Answer:
A thin layer of oil with index of refraction ng = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na= 1. A light with wavelength λ = 775 nm goes in from the air to oil and water.
Part (a) Express the wavelength of the light in the oil, λ₀, in terms of λ and n⁰ (b) Express the minimum thickness of the film that will result in destructive interference, t min, in terms of λ o
(c) Express tmin in terms of λ and no.
(d) Solve for the numerical value of tmin in nm.
Explanation:
n₀ = 1.47
refraction of water = 1.3
refraction of air = 1
wavelength λ = 775 nm
(a) wavelength of light in water ⇒ λ₀ = λ / n₀
(b) minimum thickness of the film that will result in destructive interference
t(min) = λ₀ / 2
(c) the express t(min)
t = λ /2n₀
(d) the thickness is
t = 775 / 2(1.47)
= 263.61 nm
