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.
The sleds speed when the spring returns toits uncompressed length is v = 0.03 m/s.
<u>Explanation</u>:
Given,
force constant = 42 N/cm = 0.42 N/m, mass m = 68 kg, spring x = 0.39 m
The potential energy, U, stored in the spring is
U = 1/2 kx^2
= 1 / 2
0.42
(0.39)^2
= 0.032 J
All its potential energy has been converted into kinetic energy since it has a uncompressed length.
K = 1/2 mv^2
v = sqrt (2K / m)
= √(( 2
0.032) / 68)
v = 0.03 m/s
.
E = 0.5mv^2 = 0.5*44*10^2 = 2200J
Answer:
240 ohms
Explanation:
From Ohms law we deduce that V=IR and making R the subject of the formula then R=V/I where R is resistance, I is current and V is coltage across. Substituting 120 V for V and 0.5 A for A then
R=120/0.5=240 Ohms
Alternatively, resistance is equal to voltage squared divided by watts hence 