Answer:
120 W lightbulb
Explanation:
Let the two lightbulb be A and B respectively.
Given the following data;
Power A = 120W
Power B = 90W
Voltage = 120V
To find the current flowing through each lightbulb;
a. For lightbulb A
Power = current * voltage
120 = current * 120
Current = 120/120
Current = 1 Ampere.
b. For lightbulb B
Current = power/voltage
Current = 90/120
Current = 0.75 Amperes
Therefore, the lightbulb that carries more current is A with 1 Ampere.
The speed of sound on planet is 210 m/s.
<h3>
What is Oscillation?</h3>
Oscillation is the repeating or periodic change of a quantity around a central value or between two or more states, often in time. Alternating current and a swinging pendulum are two common examples of oscillation.
There are 3 main types of Oscillation –
- Free
- damped
- forced oscillation
f = frequency = 600 Hz
lambda = wavelength = 35 cm = 0.35 m
Now,
V = speed = f × lambda = 210 m/s
Hence, speed of sound on planet is 210 m/s.
to learn more about oscillation go to -
brainly.com/question/12622728
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Because a star is a peace of the sun and the sun is hot so the star get it energy from the Sun .
Answer:
The final angular velocity is rev/s is 0.293 rev/s.
Explanation:
Given;
mass of the merry-go-round, m₁ = 120 kg
radius of the merry-go-round, r = 1.8 m
initial angular velocity, ω = 0.4 rev/s
mass of the child, m₂ = 22 kg
Apply the principle of conservation angular momentum to determine the final angular velocity;
Therefore, the final angular velocity is rev/s is 0.293 rev/s.
Answer:
D) θ₂= 36. 6º
Explanation:
In this diffraction experiment it is described by the equation
sin θ = m λ
The first dark strip occurs for m = 1 and since the angle is generally small we can approximate sine to the value of the angle
θ₁ = λ/ a
This equation is valid for linear slits, in the case of a circular slit the problem must be solved in polar coordinates, so the equation changes slightly
θ₂ = 1.22 λ / a
In the proposed exercise we start with a linear slit of width a, where tes1 = 30º and end with a circular slit of the same diameter
θ₂ = 1.22 λ / a
Let's clear (Lam/a) of equalizing the two equations
θ₁ = θ₂/ 1.22
θ₂ = 1.22 θ₁
θ₂ = 1.22 30
θ₂= 36. 6º
When reviewing the correct results is D
