The total resistance of an electric circuit with resistors widener series in the sum of the individual resistances:
Each resistor in a series circuit has a same amount of current flowing through it.
Each resistor in a parallel circuit has the same for voltage of the source applied to it.
When was this is are connected in parallel, the supply current is equal to the sum of the current through each resistor. In other words the currents in the branches of a parallel circuit add up to the supply current. When resistors are connected in parallel they have the same potential differences across them.
Answer:salt pushes up the boat or the vessle of carriage
Explanation:
<h3>
Answer:</h3>
49 N
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of the brick as 3 kg
- The coefficient of friction as 0.6
We are required to determine the force that must be applied by the woman so the brick does not fall.
- We need to importantly note that;
- For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.
- That is; Friction force = Mg
But; Friction force = μ F
Therefore;
μ F = mg
0.6 F = 3 × 9.8
0.6 F = 29.4
F = 49 N
Therefore, she must use a force of 49 N
Answer:
<h2>2 meters</h2>
Explanation:
<h2>Wavelength = Speed/Frequency </h2><h2>1000 m/s ÷ 500 hz </h2><h2>2 m</h2><h2>hz = s</h2><h2>Hopes this helps. Mark as brainlest plz!</h2>
Answer:
93.54 Hz
Explanation:
✓From the question, Number of harmonic frequency is 4
✓ the frequency (f₄ )= 116.5 Hz
✓harmonic frequency can be calculated using below expresion
fₙ = [ (nv)/4L]..........eqn(1)
v = speed of sound= 343 m/s
n = number of given harmonic frequency
L = Length of the rope
Using above expresion ,and substitute the values at (n=4) which is 4th harmonic frequency to find the " initial Lenght of the rope
fₙ = [ (nv)/4L]
f₄ = 4× 343 /4L
f₄ = 343 /L
L= 343 /f₄
But f₄= 116.5 Hz
L= 343/116.5= 2.944m
Hence, initial Lenght of the rope= 2.944m
We can determine the frequency of new length as ( initial Lenght of the rope + tubing Lenght)
= ( 2.944m + 0.721m )
= 3.667m
Hence, new length= 3.667m
To find the new frequency of the 4th harmonic we will use eqn(2)
f₄ = v/l ...............eqn(2)
From equation (2) If we substitute the values we have
f₄ = (343/3.667)
= 93.54 Hz
Hence, the the new frequency of the 4th harmonic is
93.54 Hz
