Answer:
0.53 quart
Explanation:
The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C
Since, ΔV = VγΔθ
substituting the values of the variables into the equation, we have
ΔV = VγΔθ
ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C
ΔV = 528900 × 10⁻⁶ quart
ΔV = 0.528900 quart
ΔV ≅ 0.53 quart
Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.
Answer:
The angular velocity is 15.37 rad/s
Solution:
As per the question:

Horizontal distance, x = 30.1 m
Distance of the ball from the rotation axis is its radius, R = 1.15 m
Now,
To calculate the angular velocity:
Linear velocity, v = 
v = 
v = 
v = 
Now,
The angular velocity can be calculated as:

Thus

Answer:
(a) t = 5.66 s
(b) t = 8 s
Explanation:
(a)
Here we will use 2nd equation of motion for angular motion:
θ = ωi t + (1/2)∝t²
where,
θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad
ωi = initial angular speed = 0 rad/s
t = time = ?
∝ = angular acceleration = 1.45 rad/s²
Therefore,
23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²
t² = (23.25 rad)(2)/(1.45 rad/s²)
t = √(32.06 s²)
<u>t = 5.66 s</u>
<u></u>
(b)For next 3.7 rev
θ = ωi t + (1/2)∝t²
where,
θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad
ωi = initial angular speed = 0 rad/s
t = time = ?
∝ = angular acceleration = 1.45 rad/s²
Therefore,
46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²
t² = (46.5 rad)(2)/(1.45 rad/s²)
t = √(64.13 s²)
<u>t = 8 s</u>
Frequency (f) = 500 hz (SI)
Velocity (V) = 1250 m/s (SI)
Wavelength (Lambda) = ? meters

3) Earth is about 150 million km from the Sun, and the apparent brightness of the Sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the Sun if we were located five times Earth's distance from the Sun. Answer: The Sun would appear 1/25 times as bright.