If the wagon travels 18.75 m, then the work done on the wagon is
(18.75 m) x (the steady force applied to the wagon all the way, in Newtons) .
The unit is Joules .
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
Answer: Hello the missing piece of your question is attached
question : Determine mass of steam that has entered ( in kg )
answer : 0.206 kg
Explanation:
V1 = 0.1 m^3 ,
v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg
V2 = 0.2 m^3
using the steam tables
at ; P = 1000 kPa, v' = 0.167 m^3/kg
U1 = 2321 KJ/kg
at ; P = 1000 kPa , T2 = 280°C
v'2= 0.2481 m^3kg
U2 = 2760.6
at ; P = 5MPa , T = 500°C
h1 = 3434.7 KJ/Kg
calculate final mass ( m2 )
M2 = V2 / v'2
= 0.2 / 0.2481 = 0.806 kg
therefore the mass added = m2 - m1
= 0.806 - 0.6 = 0.206 kg
i had a stroke
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Explanation:
