Answer:
D) 117 rad/s
Explanation:
We can treat this system as a circular motion where the origin is the elbow joint, the ball rotation velocity v is 35 m/s, the rotation radius is r = 0.3m.
As the ball is leaving the pitcher hand at such speed and rotation radius. Its angular velocity is:
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Answer:
b = 0.6487 kg / s
Explanation:
In an oscillatory motion, friction is proportional to speed,
fr = - b v
where b is the coefficient of friction
when solving the equation the angular velocity has the form
w² = k / m - (b / 2m)²
In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction
let's call
w₀² = k / m
w² = w₀² - b² / 4m²
b² = (w₀² -w²) 4 m²
Let's find the angular velocities
w₀² = 5 / 0.1
w₀² = 50
w = 2π f
w = 2π 1
w = 6.2832 rad / s
we subtitute
b² = (50 - 6.2832²) 4 0.1²
b = √ 0.42086
b = 0.6487 kg / s
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
Answer:
3349J/kgC
Explanation:
Questions like these are properly handled having this fact in mind;
Quantity of heat = mcΔ∅
m = mass of subatance
c = specific heat capacity
Δ∅ = change in temperature
m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)
m₁ = mass of block = 500g = 0.5kg
c₁ = specific heat capacity of unknown substance
∅₂ = block initial temperature = 50oC
∅₁ = equilibrium temperature of block and water after mix= 25oC
m₂= mass of water = 2kg
c₂ = specific heat capacity of water = 4186J/kg C
∅₃ = intial temperature of water = 20oC
0.5c₁(50-25) = 2 x 4186(25-20)
And we can find c₁ which is the unknown specific heat capacity
c₁ = 
= 3348.8J/kg C≅ 3349J/kg C
