Answer:
433 mph
Explanation:
We know that
...Eq(1)
Here,
Distance =1,592 miles
Time=3.68 hours
Putting the value of distance and Time in the Eq(1) We get
Velocity =
therefore Average velocity is 433 mph
Efficiency is the ratio of output work to input work, expressed as a percentage. Light bulbs put out less light energy than the
2 answers:
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Answer:
a) Initial speed of the ball = 14.45 m/s
b) At height 6 m speed of ball = 9.55 m/s
c) Maximum height reached = 10.65 m
Explanation:
a) We have equation of motion , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.
s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8
Substituting
Initial speed of the ball = 14.45 m/s
b) We have equation of motion , where v is the final velocity
s = 6 m, u = 14.45 m/s, a = -9.8
Substituting
So at height 6 m speed of ball = 9.55 m/s
c) We have equation of motion , where v is the final velocity
u = 14.45 m/s, v =0 , a = -9.8
Substituting
Maximum height reached = 10.65 m
Answer:
a) W =400 kJ
b) W = 0 kJ
c) W =-160.944 KJ
Explanation:
<u>Given </u>
<u><em>Process 1 ---> 2 </em></u>
The relation of the process P = constant
Pressure of point (1) P1 = 10 bar = P2
Volume of point (1) V1 = 1 m^3
Volume of point (2) V2 =4 m^3
The relation of the process V = constant
<u>Process 2 ---> 3 </u>
The relation of the process V = constant
V3 = V2
Pressure of point (3) P3 = 10 bar
Volume of point (3) V3 = 4 m^3
<u>Process 3 ---> 1 </u>
The relation of the process PV = constant
<u>Required </u>
Sketch the processes on the PV coordinates
The work for each process in kJ
<u>Solution </u>
The work is defined by
W=
<em>a=V2</em>
<em>b=V1</em>
<em>x=P</em>
<em>dx=dV</em>
<u>Process 1 ---> 2 </u>
P3 = P4 = 5 bar
W=
<em>a=V3</em>
<em>b=V2</em>
<em>x=4</em>
<em>dx=dV</em>
putting the value of a, b, x, dx in above integral
W=400 kJ
<u>Process 2 ---> 3 </u>
V = constant Then there is no change in the volume,hence W = 0 kJ
<u>Process 3 ---> 1 </u>
By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1
W=
a=V1
b=V3
x=1V^-1
dx=dV
putting the value of a, b, x, dx in above integral
W=| ln V | limit a and b
= -160.944 KJ
