If we increase the length of the resistor, resistance will also increase and vice-versa.
It is given that in a wire simulation, we have to adjust only the length of the resistor.
It required to find the effect on resistor when adjust only the length of the resistor.
<h2>What happens to resistor when adjust only the length of the resistor?</h2>
As we know that the length is directly proportional to the resistance of the wire that means is we increase the size or length of the wire then it also increase the resistance present in the wire. On the other hand if we decrease the length of the wire, it will decrease the resistance. This can be represented as:
Resistance = resistivity × length/area
This is also because there is more collision of the flowing electrons in the wire to the metal ions present there.
Thus if we increase the length of the resistor, resistance will also increase and vice-versa.
Learn more about the resistance here:
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Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Answer:
d y / x = m λ
Explanation:
When the laser beam, which is a coherent light, hits the slits, part of each beam passes through each slit.
When this is observed on a screen that is quite far from the slits, a series of intense linear separated by dark areas. The explanation for this distribution of the light pattern is that when adding the rays that come out of the slits they travel different distances, which introduces a difference in optical path and if this difference is an integer multiple of the wavelength, a bright line
d sin θ = m λ
Where d is the distance between the slits (0.1 mm)
Also, since the angle of the measurements is small, we can approximate the tangent
tan θ = y / x = sin θ /sin θ
sint θ = y / x
Substituting into the equation
d y / x = m λ
This expression gives the location of the bright lines on the screen
Explanation:
A.
Using the wave equation:
v = d/t
Where,
v = velocity
d = distance
= 23.5 km
= 23500 m
t = time
= 2m 30s = 150 s
v = 23500/150
= 156.7 m/s
B.
From the calculated value above, it can be seen that the velocity of the heat blast is >>> than the velocity of sound above sea level.
Answer:
45C in a minute is
45/60 C in a second
.75 C in a second is 3/4 of an ampere.
Explanation: