Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
The drawbar or other connections must be strong enough to pull all the weight of the vehicle being towed. The drawbar or other connection may not exceed 15 feet from one vehicle to the other.
Question
Determine the average water exit velocity
Answer:
53.05 m/s
Explanation:
Given information
Volume flow rate,
Diameter d= 8cm= 0.08 m
Assumptions
- The flow is jet flow hence momentum-flux correction factor is unity
- Gravitational force is not considered
- The flow is steady, frictionless and incompressible
- Water is discharged to the atmosphere hence pressure is ignored
We know that Q=AV and making v the subject then
where V is the exit velocity and A is area
Area, where d is the diameter
By substitution
To convert v to m/s from m/s, we simply divide it by 60 hence
Answer:
Carpenter's square
Explanation:
The most common hand tool used to measure or set angles with its application extending to setting angles of roofs and rafters. Another name of a Carpenter's square is a framing square.
Other hand tools that are used to measure angles are;
- The combination square that allows a user to set both 90° and 45° angles
- A Bevel that allows users to set any angle they like.
- A Protractor that resembles a bevel but its marks are marked in an arc.
- An electromagnetic angle finder which gives a reading according to the measure of the arms adjusted by the user.
Answer:
5984.67N
Explanation:
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?
from continuity equation
v1A1=v2A2
equation of continuity
v1=4ft /s=1.21m/s
d1=14 inch=.35m
d2=14-2=0.304m
A1=pi*d^2/4
0.096m^2
a2=0.0706m^2
from continuity once again
1.21*0.096=v2(0.07)
v2=1.65
force on the pipe
(p1A1- p2A2) + m(v2 – v1)
from bernoulli
p1 + ρv1^2/2 = p2 + ρv2^2/2
difference in pressure or pressure drop
p1-p2=2psi
13.789N/m^2=rho(1.65^2-1.21^2)/2
rho=21.91kg/m^3
since the pipe is cylindrical
pressure is egh
13.789=21.91*9.81*h
length of the pipe is
0.064m
AH=volume of the pipe(area *h)
the mass =rho*A*H
0.064*0.07*21.91
m=0.098kg
(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)
force =5984.67N