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

- Speed of the mobile = 250 m/s
- It starts decelerating at a rate of 3 m/s²
- Time travelled = 45s

- Velocity of mobile after 45 seconds

We can solve the above question using the three equations of motion which are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

We are provided with,
- u = 250 m/s
- a = -3 m/s²
- t = 45 s
By using 1st equation of motion,
⇛ v = u + at
⇛ v = 250 + (-3)45
⇛ v = 250 - 135 m/s
⇛ v = 115 m/s
✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
The answer is 100mm/s. I hope this helps :)
Answer:
2 kg
Explanation:
Acceleration = 5 m/s^2
Force = 10 N
Force = mass * acceleration
mass = force / acceleration
mass = 10 / 5
mass = 2 kg
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