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:
Explanation:
This problem bothers on the energy stored in a spring in relation to conservation of energy
Given data
Mass of block m =200g
To kg= 200/1000= 0.2kg
Spring constant k = 1.4kN/m
=1400N/m
Compression x= 10cm
In meter x=10/100 = 0.1m
Using energy considerations or energy conservation principles
The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring
Potential Energy stored in spring
P.E=1/2kx^2
Kinetic energy of the block
K.E =1/mv^2
Where v = velocity of the block
K.E=P.E (energy consideration)
1/2kx^2=1/mv^2
Kx^2= mv^2
Solving for v we have
v^2= (kx^2)/m
v^2= (1400*0.1^2)/0.2
v^2= (14)/0.2
v^2= 70
v= √70
v= 8.36m/s
a. Distance moved if the ramp exerts no force on the block
Is
S= v^2/2gsinθ
Assuming g= 9. 81m/s^2
S= (8.36)^2/2*9.81*sin60
S= 69.88/19.62*0.866
S= 69.88/16.99
S= 4.11m
We don't know. A black hole is a star that has collapsed into its own gravity. The gravity in fact, is so strong that even light cannot get through it. That's why it looks black to us.
