Answer:
I need this for may schooling
Answer:
a) 0.1832 A
b) 11.91 Volts
c) 2.18 Watt , 0.0168 Watt
Explanation:
(a)
R = external resistor connected to the terminals of the battery = 65 Ω
E = Emf of the battery = 12.0 Volts
r = internal resistance of the battery = 0.5 Ω
i = current flowing in the circuit
Using ohm's law
E = i (R + r)
12 = i (65 + 0.5)
i = 0.1832 A
(b)
Terminal voltage is given as
= i R
= (0.1832) (65)
= 11.91 Volts
(c)
Power dissipated in the resister R is given as
= i²R
= (0.1832)²(65)
= 2.18 Watt
Power dissipated in the internal resistance is given as
= i²r
= (0.1832)²(0.5)
= 0.0168 Watt
The answer is B, or endoplasmic reticulum.
Answer:
141 m at 65.6° N of E
Explanation:
Let E be along the positive x axis of a unit circle
N = 90°
E = 0°
SE = -45°
W = 180°
NW = 135°
east displacement
x = 140cos90 + 85cos0 + 35cos-45 + 38cos180 + 19cos135 = 58.313708... m
north displacement
y = 140sin90 + 85sin0 + 35sin-45 + 38sin180 + 19sin135 = 128.6862915... m
d = √(128.6862915² + 58.313708²) = 141.28216525... m
tanθ = 128.6862915 / 58.313708
θ = 65.622521...