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mart [117]
3 years ago
8

10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave

pulse moves from one end of the rope to another and find that it is 36 m/s. What frequency must they vibrate the rope at to create the second harmonic
Physics
1 answer:
zhannawk [14.2K]3 years ago
5 0

Answer:

To create a second harmonic the rope must vibrate at the frequency of 3 Hz

Explanation:

First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,

f₁ = v/2L

where,

v = speed of wave = 36 m/s

L = Length of rope = 12 m

f₁ = fundamental frequency

Therefore,

f₁ = (36 m/s)/2(12 m)

f₁ = 1.5 Hz

Now the frequency of nth harmonic is given in general, as:

fn = nf₁

where,

fn = frequency of nth harmonic

n = No. of Harmonic = 2

f₁ = fundamental frequency = 1.5 Hz

Therefore,

f₂ = (2)(1.5 Hz)

<u>f₂ = 3 Hz</u>

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amid [387]

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F

Then, it means that the net force acting on the test charge has a magnitude of √2F.

7 0
3 years ago
A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure
kozerog [31]

Answer:

43.7 °C

Explanation:

\alpha_b = Coefficient of linear expansion of brass = 18\times 10^{-6}\ ^{\circ}C

\alpha_s = Coefficient of linear expansion of steel = 11\times 10^{-6}\ ^{\circ}C

L_{0b} = Initial length of brass = 31 cm

L_{0s} = Initial length of steel = 11 m

\Delta L = Total change in length = 3 mm

Total change in length would be

\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C

\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C

The final temperature is 43.7 °C

6 0
3 years ago
Lowest frequency of wave
Harlamova29_29 [7]
The answer would be a radio wave
4 0
3 years ago
Read 2 more answers
1. Impulse is the product of force and distance. True or false?
daser333 [38]

Answer:

Explanation:

1. Impulse, I = F.t

  The statement impulse is the product of Force and distance is false.

2. F = m g

   Force necessary to lift the object depends on the mass of the object.

   statement 2 is false.

3. Joule is equal to Newton times meter.

    Statement 3 is false.

4. Work done to lift an object is correct statement.

   Statement 4 is true.

5. Kinetic energy of an object is due to motion.

  Statement 5 is false.

6. Stopping distance is directly proportional to the square of velocity.

     If velocity is doubled, stopping distance is quadrupled.

    Statement 6 is false.

8 0
3 years ago
The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
2 years ago
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