A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
1 answer:
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ = ................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
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Answer:
T = 858.25 s
Explanation:
Given data:
Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),
initial uniform temperature ( Ti ) = 200 c
Final temperature = 550 c
convection coefficient = 250 w/m^2 k
products combustion temp = 800 c
calculate how long the plate should be left in the furnace ( to attain 550 c )
first calculate/determine the Fourier series Number ( Fo )
= 0.4167 =
therefore Fo = 3.8264
Now determine how long the plate should be left in the furnace
Fo =
k = 48
p = 7830
L = 0.1
Input the values into the relation and make t subject of the formula
hence t = 858.25 s
Answer:
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Explanation:
Given that
Heat loss from room (Q)= 60 KJ/min
Work supplied to the room(W) = 1.2 KW = 1.2 KJ/s
We know that 1 W = 1 J/s
Sign convention for heat and work:
1. If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.
2. If work is done by the system then it is taken as positive and if work is done on the system then it is taken as negative.
So
Q = -60 KJ/min
In 10 min Q = -600 KJ
W = -1.2 KJ/s
We know that
1 min = 60 s
10 min = 600 s
So W = -1.2 x 600 KJ
W = -720 KJ
WE know that ,first law of thermodynamics
Q = W + ΔU
-600 = - 720 + ΔU
ΔU = 120 KJ
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Answer:
Examples of reciprocating motion in daily life are;
1) The needles of a sewing machine
2) Electric powered reciprocating saw blade
3) The motion of a manual tire pump
Explanation:
A reciprocating motion is a motion that consists of motion of a part in an upward and downwards or in a backward and forward (↔) direction repetitively
Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump
In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.