Answer:
a) 2nd case rate of rotation gives the greater speed for the ball
b) 1534.98 m/s^2
c) 1515.04 m/s^2
Explanation:
(a) v = ωR
when R = 0.60, ω = 8.05×2π
v = 0.60×8.05×2π = 30.34 m/s
Now in 2nd case
when R = 0.90, ω = 6.53×2π
v = 0.90×6.53×2π = 36.92 m/s
6.35 rev/s gives greater speed for the ball.
(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2
(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2
Answer:
The value of the average convection coefficient is 20 W/Km².
Explanation:
Given that,
For first object,
Characteristic length = 0.5 m
Surface temperature = 400 K
Atmospheric temperature = 300 K
Velocity = 25 m/s
Air velocity = 5 m/s
Characteristic length of second object = 2.5 m
We have same shape and density of both objects so the reynold number will be same,
We need to calculate the value of the average convection coefficient
Using formula of reynold number for both objects



Here, 


Put the value into the formula


Hence, The value of the average convection coefficient is 20 W/Km².
The correct answer would be 1.375 < t < 3 i hope this helps anyone
Answer:
The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.
The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds
Answer:
refractive index of the unknown material is 1.33.
Explanation:
μ₁ = 1.21
incidence angle (i) = 41.9°
refraction angle (r) = 37.3°
Let us assume μ be the refractive index of the unknown material
according to snell's law of refraction.
μ₁ sin i = μ₂ sin r
1.21 × sin 41.9° = μ × sin 37.3°
μ = 1.33
hence the refractive index of the unknown material comes out top be 1.33