Point of reference - an indicator that orients you generally; "it is used as a reference for comparing the heating and the electrical energy involved"
3.4814815 (or 3 13/27) m/s
speed = distance/time
3.4814815 (or 3 13/27) = 94/27
Answer
given,
constant speed of cart on right side = 2 m/s
diameter of nozzle = 50 mm = 0.05 m
discharge flow through nozzle = 0.04 m³
One-fourth of the discharge flows down the incline
three-fourths flows up the incline
Power = ?
Normal force i.e. Fn acting on the cart
v is the velocity of jet
Q = A V
v = 20.37 m/s
u be the speed of cart assuming it to be u = 2 m/s
angle angle of inclination be 60°
now,
F n = 2295 N
now force along x direction
Power of the cart
P = F x v
P = 1987.52 x 20.37
P = 40485 watt
P = 40.5 kW
Complete question :
A 12 m x 15 m house is built on a 12-cm-thick concrete slab.
What is the heat-loss rate through the slab if the ground temperature is 5°C while the interior of the house is 25°C
Answer:
3kW
Explanation:
Given the following :
Dimension of house :
Length = 12m
Width = 15m
Thickness of concrete slab (t) = 12cm
t in metres :
100cm = 1m
12cm = (12/100)m
= 0.12m
Ground temperature (Tg) = 5°C
Interior temperature = (Th) = 25°C
Thermal conductivity of concrete (K) is approximately 1 Wm/k
Using the relation:
Q = KA * [ (Th - Tg) / d]
A = Length * width = (12 *15) = 180
Q = (1 * 180) * [(25°C - 5°C) / 0.12]
Q = 180 * (20/0.12)
Q = 180 * 16.6666
Q = 3,000W = 3kW
12 MPH
I DIDNT do the math my brother did hes in colledge so good lucks guys
