True will be your answer have a great day
Answer:
b. Discharging; anode; cathode
Explanation:
When discharging , it means the battery is producing a flow electric current, the lithium ions are released from the anode to the cathode which generates the flow of electrons from one side to another. When charging Lithium ions are released by the cathode and received by the anode.
C both A and b cause they are technician both technicians so they both measure out the floor pan reinforcement be designed to transfer collision energy so I say both A and B
Answer:
The distance measure from the wall = 36ft
Explanation:
Given Data:
w = 10
g =32.2ft/s²
x = 2
Using the principle of work and energy,
T₁ +∑U₁-₂ = T₂
0 + 1/2kx² -wh = 1/2 w/g V²
Substituting, we have
0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²
170 = 0.15528V²
V² = 170/0.15528
V² = 1094.796
V = √1094.796
V = 33.09 ft/s
But tan ∅ = 3/4
∅ = tan⁻¹3/4
= 36.87°
From uniform acceleration,
S = S₀ + ut + 1/2gt²
It can be written as
S = S₀ + Vsin∅*t + 1/2gt²
Substituting, we have
0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)
19.85t - 16.1t² + 3 = 0
16.1t² - 19.85t - 3 = 0
Solving it quadratically, we obtain t = 1.36s
The distance measure from the wall is given by the formula
d = VCos∅*t
Substituting, we have
d = 33.09 * cos 36. 87 * 1.36
d = 36ft
Answer:
![V_{dc}=84.15\ V](https://tex.z-dn.net/?f=V_%7Bdc%7D%3D84.15%5C%20V)
Explanation:
Given that
Vrms= 60 V
Vf= 0.7 V
We know that peak value of AC voltage given as
![V_{o}=\sqrt{2}\ V_{rms}](https://tex.z-dn.net/?f=V_%7Bo%7D%3D%5Csqrt%7B2%7D%5C%20V_%7Brms%7D)
Now by putting the values
![V_{o}=60\sqrt{2}\ V](https://tex.z-dn.net/?f=V_%7Bo%7D%3D60%5Csqrt%7B2%7D%5C%20V)
The output voltage of the DC current given as
![V_{dc}=V_{o}-V_f](https://tex.z-dn.net/?f=V_%7Bdc%7D%3DV_%7Bo%7D-V_f)
![V_{dc}=60\sqrt{2}-0.7\ V](https://tex.z-dn.net/?f=V_%7Bdc%7D%3D60%5Csqrt%7B2%7D-0.7%5C%20V)
![V_{dc}=84.15\ V](https://tex.z-dn.net/?f=V_%7Bdc%7D%3D84.15%5C%20V)
Therefore output voltage of the DC current is 84.15 V.