Answer:
i dont know
Explanation:
but what you can do is ask you mom or dad to get you a tutor to help you
Answer
given,
resistance = 0.05 Ω
internal resistance of battery = 0.01 Ω
electromotive force = 12 V
a) ohm's law
V = IR
and volage
now,

inserting the values
I = 200 A
b) Voltage
V = I R
V = 200 x 0.05
V = 10 V
c) Power
P = I V
P = 200 x 10 = 2000 W
d) total resistance = 0.05 + 0.09 = 0.14 Ω
I = 80 A
V = 80 x 0.05 = 4 V
P = 4 x 80 = 320 W
Answer:
Goal or Field Goal
Explanation:
It is a goal in a sport like hockey or it is a field goal in football.
Answer:
a) θ₁ = 23.14 °
, b) θ₂ = 51.81 °
Explanation:
An address network is described by the expression
d sin θ = m λ
Where is the distance between lines, λ is the wavelength and m is the order of the spectrum
The distance between one lines, we can find used a rule of proportions
d = 1/600
d = 1.67 10⁻³ mm
d = 1-67 10⁻³ m
Let's calculate the angle
sin θ = m λ / d
θ = sin⁻¹ (m λ / d)
First order
θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)
θ₁ = sin⁻¹ (3.93 10⁻¹)
θ₁ = 23.14 °
Second order
θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)
θ₂ = sin⁻¹ (0.786)
θ₂ = 51.81 °
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC