All categories

3 years ago

Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On

Physics

1 answer:

8090 [49]3 years ago

4 0

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.

You might be interested in

A cube of plastic 1.2x10^-5 km on a side has a mas of 1.1 g. what is its density in g/cm^3?

Mamont248 [21]

Density = mass / volume

mass = 1.1 g

volume = length of side ^ 3 = [1.2 * 10^-5 km * 100000 cm/km]^3 = [1.2 cm]^3 = 1.728 cm^3

density = 1.1 g / 1.728 cm^3 = 0.64 g / cm^3

5 0

4 years ago

A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.

ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

$P = VI$

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

$I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\$

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

$V = IR$

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

$R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega$

Therefore, the resistance of the bulb is 484 Ω

3 0

3 years ago

Read 2 more answers

What is the momentum of a 73 person running with a speed of 2 m/s

NemiM [27]

Answer:

Explanation:

Idk

6 0

4 years ago

Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0

ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

$F_c=m\frac{v^2}{r}$ (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

$v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}$

you replace this value of v in the equation (1). Also, you replace the values of r and m:

$F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N$

hence, the tension in the string must be T = Fc = 764.41 N

5 0

3 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

Explanation:

Given that,

Current = 40 A

Magnetic field $B=3.7\times10^{-5}\ T$

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

$B'=\dfrac{\mu_{0}I}{2\pi r}$

Where, r = distance

I = current

Put the value into the formula

$B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}$

$B'=1.8\times10^{-5}\ T$

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

$B''=\sqrt{B^2+B'^2}$

Put the value into the formula

$B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}$

$B''=4.11\times10^{-5}\ T$

Hence, The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

6 0

3 years ago