The direction of applied force has to be .............. to the distance in order to say work is done
1 answer:
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Answer:
the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
Explanation:
The computation of the magnitude of the each force is shown below:
Provided that
Ratio of forces = 3: 5
Let us assume the common factor is x
Now
first force = 3x
And, the second force = 5x
Resultant force = 35 N
The Angle between the forces = 60 degrees
Based on the above information
Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos
35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]
35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)
35 = √49 x²
x = 5
So, the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
Answer:
A. The work done during the process is W = 0
B. The value of heat transfer during the process Q = 442.83
Explanation:
Given Data
Initial pressure = 100 k pa
Initial temperature = 25 degree Celsius = 298 Kelvin
Final pressure = 300 k pa
Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.
⇒ =
------------- (1)
Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.
⇒ P ∝ T
⇒ =
⇒ Put all the values in the above formula we get the final temperature
⇒ =
× 298
⇒ = 894 Kelvin
(A). Work done during the process is given by W = P × ()
From equation (1), =
so work done W = P × 0 = 0
⇒ W = 0
Therefore the work done during the process is zero.
Heat transfer during the process is given by the formula Q = m (
-
)
Where m = mass of the gas = 1 kg
= specific heat at constant volume of nitrogen = 0.743
Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )
⇒ Q = 442.83
Therefore the value of heat transfer during the process Q = 442.83
Answer:
y(i) = h
v(y.i) = 0
Explanation:
See attachment for elaboration
