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alexira [117]
3 years ago
8

An air-filled pipe is found to have successive harmonics at 945 Hz , 1215 Hz , and 1485 Hz . It is unknown whether harmonics bel

ow 945 Hz and above 1485 Hz exist in the pipe. What is the length of the pipe

Physics
1 answer:
Aleonysh [2.5K]3 years ago
8 0

Answer:

L = 0.635m

Explanation:

This problem involves the concept of stationary waves in pipes. For pipes closed at one end,

The frequency f = nv/4L for n = 1,3,5....n

For pipes open at both ends

f = nv/2L for n = 1,2,3,4...n

Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.

The film solution can be found in the attachment below.

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A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s
skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I =  0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

6 0
3 years ago
A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m
Irina18 [472]
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂'  --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' =  -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
7 0
3 years ago
A transformer is designed to provide 6 V from a 150 V supply. If the primary coil has 1000 turns, how many turns does the second
zavuch27 [327]
The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.

Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.

Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
7 0
3 years ago
How much work is accomplished when a force of 250 N pushes a box across the floor for a distance of 50 meters?
IrinaK [193]
Work = force * distance
and newton*meters = Joule

In this case,  work = 250N*50m = 12500 J

So the answer is D) 12,500 J

3 0
4 years ago
A table tennis ball has a mass of about 2.45 g. Suppose the ball is hit across the table with a speed of 4.0 m/s. What is it’s k
cluponka [151]

Answer:

0.0196 j

Explanation:

i) The formula for kinetic energy is as follows: 0.5*m*v^2

ii) Since we have all the values all that's left is to plug them into the equation

iii) First, WE MUST, Convert grams into kgs as this is the SI unit of mass so 2.45/1000

iv) All that's left now is to plug it into the equation so:

0.5* (s.45/1000)*(4^2)

v) Lastly we add the unit joules at the end as we're talking about energy

Hope this was useful! :)

3 0
2 years ago
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