When two or more resistors are connected in parallel to a battery A) the voltage across each resistor is the same. B) the total
1 answer:
Answer:
Here, for parallel resisitors
Option D) all of the other choices are are true
is correct.
Explanation:
In parallel connection:
1) Voltage across each element connected in parallel remain same.
2) Kirchhoff's Current Law (KCL), sum of the current entering and leaving the junction will be zero
3) The equivalent resistance of the elements connected in parallel is always less than the individual resistance of any resistor in the circuit.
The equivalent resistance in a parallel circuit is given by:
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Answer:
λ = 482.05 nm
Explanation:
The diffraction phenomenon and the diffraction grating is described by the expression
d sin θ = m λ
where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction
in this case they indicate the distance between slits, the angle and the order of diffraction
λ = d sin θ / m
let's calculate
λ = 1.00 10⁻⁶ sin 74.6 / 2
λ = 4.82048 10⁻⁷ m
Let's reduce to nm
λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)
λ = 482.05 nm
Answer:
η = 1.31
Explanation:
The formula for the refractive index of from air to some other medium is given by the following formula:
where,
η = refractive index = ?
c = speed of light in air = 3 x 10⁸ m/s
v = speed of light in ice = 2.29 x 10⁸ m/s
Therefore, using these values in the equation we get:
<u>η = 1.31</u>
Answer:
hello your question has some missing values attached below is the complete question with the missing values
answer :
a) 0.083 secs
b) 0.33 secs
c) 3e^-4/3
Explanation:
Given that
g = 32 ft/s^2 , spring constant ( k ) = 2 Ib/ft
initial displacement = 1 ft above equilibrium
mass = weight / g = 4/32 = 1/8
damping force = instanteous velocity hence β = 1
a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>
time mass passes through equilibrium = 1/12 seconds = 0.083
<u>b) Calculate the time at which the mass attains its extreme displacement </u>
time when mass attains extreme displacement = 1/3 seconds = 0.33 secs
<u>c) What is the position of the mass at this instant</u>
position = 3e^-4/3
attached below is the detailed solution to the given problem
