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
Answer:
Y component = 32.37
Explanation:
Given:
Angle of projection of the rocket is, 
Initial velocity of the rocket is, 
A vector at an angle 
with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.
If a vector 'A' makes angle with the horizontal, then the horizontal and vertical components are given as:
Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:
Plug in the given values and solve for 
. This gives,
Therefore, the Y component of initial velocity is 32.37.
Where the answer choice then i can answer it
Answer:
A) 11.28 x 10^(7) A.m²
B) 2.258 x 10^(17)A
Explanation:
A) The current density is given by the formula ;
J = nqv
Where n is the density of protons in the solar wind which is 12.5 cm³ or 12.5 x 10^(-6) m³
q is the proton charge which is 1.6 x 10^(-19) C
v is velocity which is 564km or 564000m
Thus, J = 12.5 x 10^(-6) x 1.6 x 10^(-19) x 564000 = 11.28 x 10^(7) A.m²
B) the formula for the total current the earth received is given as;
I = JA
The effective area is the cross section of the earth and thus,
Area = πr² where r is the radius of the earth given as: 6.371 x 10^(6)
A = π(6.371 x 10^(6)) ²
So I = 11.28 x 10^(7) x π(6.371 x 10^(6))² = 2.258 x 10^(17)A
