Answer:
Total tension in the person's arm = 58.82 N
Explanation:
Normal Contact Force: Normal contact force is the vertical component of the contact force that is perpendicular to the surface that an object makes contact with.
As shown in fig.1 in the diagram on the attached file below.
The Normal contact force (R) = Tsinθ = The vertical component of T
Where T = Total tension in the person's arm (N), θ = angle of inclination to the horizontal.
∴ R = Tsinθ
T = R/sinθ where R= 23N, θ =23°
T = 23/sin23° = 23/0.391
T = 58.82 N
∴ Total tension in the person's arm = 58.52 N
Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as
FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so
Change in entropy
=3( 3.36035 - 2.88) = 1.44 kJ/kg
Answer:
W =1562.53 N
Explanation:
It is given that,
Radius of the aluminium ball, r = 24 cm = 0.24 m
The density of Aluminium,
We need to find the thrust and the force. The mass of the liquid displaced is given by :
V is volume
Weight of the displaced liquid
W = mg
So,
So, the thrust and the force is 1562.53 N.
Answer: Option (b) is the correct answer.
Explanation:
Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.
Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.
Thus, we can conclude that negative charge spread evenly on both ends.
Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is <em>PV</em> = <em>C</em>
<em>T = C </em> for isothermal process
Calculating for the work done in isothermal process
<em>W</em> = <em>P</em>₁<em>V</em>₁
= <em>mRT</em>₁ [∵<em>pV</em> = <em>mRT</em>]
= (5) (0.287) (272.039)
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
Δ<em>U</em> =
From 1st law of thermodynamics
Q = Δ<em>U </em>+ <em>W</em>
= 0 + 270.588
= 270.588 kJ