The bell rings and a physics student heads to class. They stop to talk to a few friends. They slow
1 answer:
Answer:
420m
Explanation:
Given parameters:
Time = 5minutes
Average speed = 1.4m/s
Unknown:
Distance covered = ?
Solution:
Speed is the rate of change of distance with time.
Mathematically;
Speed =
Distance = speed x time
Insert the parameters and solve;
Convert the time to seconds;
1 minute = 60s
5 minute = 5 x 60 = 300s
So,
Insert the parameters and find the distance;
Distance = 300 x 1.4 = 420m
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Answer:
∑F = 10.2 N
Explanation:
We have:
Initial velocity: 0.5 m/s
Final velocity: 3 m/s
Time: 1.5 s
We have all of the components needed to calculate acceleration. Let's do that, shall we?
a = vf-vo/t
a = 2.5/1.5
a = 1.7 /
Now, look at the Net Force equation:
∑F = ma
Plug in the variables, to get:
∑F = (6)(1.7)
∑F = 10.2 N (You can round this according to significant digits)
Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v + xv
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v + x₂v
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ