Answer:
u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)
Explanation:
Given:
- Electric Field strength near earth's surface E = 145 V / m
- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg
Find:
- How much energy is stored per cubic meter in this field?
Solution:
- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:
u_e = 0.5*e_o * E^2
- Plug in the values given:
u_e = 0.5*8.854 *10^-12 *145^2
u_e = 9.30777 * 10^-8 J/m^3
The available motor sizes for 2023 Ariya AC synchronous drive motor systems are:
40 kW.
62 kW.
160 kW.
<h3>What is a synchronous motor?</h3>
A synchronous motor refers to an alternating current (AC) electric motor in which the rotational speed of the shaft is directly proportional (equal) to the frequency of the supply current, especially at a steady state.
In Engineering, the available motor sizes for 2023 Nissan Ariya AC synchronous drive motor systems include the following:
40 kW.
62 kW.
160 kW.
Read more on synchronous motor here:
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Answer:
True
Explanation:
A semicircular or circular torch movement should be used when depositing weld beads.
Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt

- Integrate and evaluate the on the interval:

- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:

- Integrate and evaluate the on the interval:

Answer:
The percentage of the remaining alloy would become solid is 20%
Explanation:
Melting point of Cu = 1085°C
Melting point of Ni = 1455°C
At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.
The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:
60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy
At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:
40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.
The percentage of remaining alloy in solid is equal to
Solid = (10/50) * 100 = 20%