Answer:
The work done against gravity is 78.4 J
Explanation:
The work is calculated by multiplying the force by the distance that the
object moves
W = F × d, where W is the work , F is the force and d is the distance
The SI unit of work is the joule (J)
We need to find the work done against gravity when lowering a
16 kg box 0.50 m
→ F = mg
→ m = 16 kg, and g = 9.8 m/s²
Substitute these value in the rule
→ F = 16 × 9.8 = 156.8 N
→ W = F × d
→ F = 156.8 N and d = 0.50
Substitute these values in the rule
→ W = 78.4 J
<em>The work done against gravity is 78.4 J</em>
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
Explanation:
The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.
The relation between wave period and frequency is as follows.
T = \frac{1}{f}T=
f
1
where, T = time period
f = frequency
It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.
T = \frac{1}{f}T=
f
1
or, f = \frac{1}{T}f=
T
1
= \frac{1}{18 sec}
18sec
1
= 0.055 per second (1cycle per second = 1 Hertz)
or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz
<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>
