A 6.41 $\mu C$ particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive $x$
1 answer:
Answer:
The particle's velocity is 212.15 m/s.
Explanation:
Given that,
Charge of particle,
The magnitude of electric field, E = 1270 N/C
The magnitude of magnetic field, B = 1.28 T
Net force,
We need to find the magnitude of the particle's velocity. the net force acting on the particle is given by Lorentz force as :
So, the particle's velocity is 212.15 m/s.
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Answer:
If I double the current in the inductor, the new total energy will become 4E (option f).
Explanation:
The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:
where E is Energy [J], L is Inductance [H] and I is Current [A].
If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:
Then:
<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>
Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus, =
= 171.82 kJ/kg
We know COP of heat pump
COP =
=
= 6.076
Therefore, Work out put, W =
= 171.82 / 6.076
= 28.27 kJ/kg