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Consider a telescope with a small circular aperture of diameter 2.0 centimeters.
If two point sources of light are being imaged by this telescope, what is the maximum wavelength (lambda) at which the two can be resolved if their angular separation is 3.0x10^-5 radians? Express your answer in nanometers
Answer:
λ =492nm
Explanation:
Given Data
diameter=2.0 cm
angular separation=3.0x10^-5 radians
λ=?
Solution
sin(theta) = 1.22 x λ /D
λ=(0.02×sin(3.0×10⁻⁵))/1.22
λ=492nm
Answer:
<h3>
FOR PARALLEL CONNECTION</h3><h3>I1 = 0.12A</h3><h3>I2 = 0.12A </h3><h3>IT =0.24A</h3><h3>FOR SERIES CONNECTION</h3><h3>I1 = I2 = 0.06A</h3><h3>IT =0.06A</h3><h3 />
Explanation:
According to ohms law, V =ITRt
V is the supply voltage
IT is the total current flowing in the circuit
Rt is the total equivalent resistance
Given R1= R2= 100Ω
V= 12V
FOR PARALLEL CONNECTION;
To calculate the total current IT in the battery, we need to calculate the total equivalent resistance RT first. For a parallel connected circuit, the equivalent resistance in the circuit is the sum of the reciprocal of its individual resistances as shown;
RT = 50Ω
from the equation above;
IT = V/RT
IT = 12/50
IT = 0.24A
Note that in a parallel connected circuit, different current flows through the resistances but the same voltage is across them.
IT = I1+I2
For current in resistance R1;
I1 = V/R1
I1 = 12/100
I1 = 0.12A
Since both resitance are the same, they will share the total current equally. Therefore I2 = 0.12A
FOR SERIES CONNECTION;
The total equivalent resistance in the circuit will be the sum of their individual resistances.
RT = 100Ω+100Ω
RT = 200Ω
IT = V/RT
IT = 12/200
IT = 0.06A
Since the resistances are connected in series, the same current will flow through them but different voltages. The total current flowing in the circuit will be the same current flowing through the resistors.
Therefore I1 = I2 = 0.06A
Answer:
4.96 km/hr
Explanation:
Given that,
A person is running west in Moscow at 2.1 km/hr relative to the ground
Another person is jogging east at 4.5 km/hr.
We need to find the velocity of the jogger relative to the ground. Using the concept of relative speed, let V is the required velocity. So,
So, the velocity of the jogger relative to the ground is 4.96 km/hr.