Answer:
Mass of ice per second melt is 2.74×10^-5Kg/s
Explanation:
Temperature of one end of the copper rod is 100°C boiling point of water and the other end of the rod is 0°C
Temperature difference in the copper rod = 100 - 0 = 100°C
Cross sectional area = 3.6×10^-4m^2
Length of rod , L = 1.7m
Amount of heat transfer from the boiling water to the ice water mix through the copper rod is given by:
Q = KA◇T/ L
Q = (390×(3.6×10^-4)×100°C)/1.7
Q = 14.04/1.7
Q = 8.26J/s
From the equation
Q = mLf
m = Q/ Lf
Where Lf = Latent heat of fusion for water= 3.34×10^5J/Kg
m = 8.26/(3.34×10^5)
m = 2.74×10^-5Kg/s
Answer:
a circuit that is a blend of series paths and parallel paths. See Figure for a visual explanation. Most circuits are of this variety. Don't be afraid to tackle these circuits as far as the math goes.
Answer:
1299 N/m²
Explanation:
Applying,
P = ρgh............... Equation 1
Where P = pressure of water at the bottom of the container, ρ = density of water, h = height of water in the container, g = acceleration due to gravity.
but,
V = πr²h............ Equation 2
Where V = volume of water, r = radius of the container
make h the subject of the equation
h = V/πr².................. Equation 3
From the question,
Given: V = 200 liters, = 200000 cm³, r = 70 cm
Constant: π = 22/7
Substitute these values into equation 3
h = 200000/[(22/7)(70²)]
h = 12.99 cm
h = 0.1299 m
Also given
Constant: g = 10 m/s², ρ = 1000 kg/m³
Substitute these values into equation 1
P = 0.1299(10)(1000)
p = 1299 N/m²
Answer:
2.5 kg
Explanation:
We can solve the problem by using Newton's second law:
where
F is the net force acting on the train
m is the mass of the train
a is the acceleration
For the toy train in the problem,
F = 3.0 N
a = 1.2 m/s^2
So we can solve the formula for m, to find the mass of the train:
