Ask question

Ask question

All categories

2 years ago

13

Which of the following quantities provide enough information to calculate the tension in a string of mass per unit length μ that

vibrates with a frequency f?A. the wave speed v and ?
B. the wavelength ? and f
C. the wave speed v and f
D. the wavelength ? and ?

1 answer:

Bad White [126]2 years ago

8 0

Answer:

A. the wave speed v and Wavelength

Explanation:

Given that

Mass density per unit length=μ

Frequency = f

The velocity V given as

$\mu=\dfrac{T}{V^2}\ kg/m$

$V=\sqrt{\dfrac{T}{\mu}}$

T=Tension

V=Velocity

V= f λ

λ=Wavelength

Therefore to find the tension ,only wavelength and speed is required.

The answer is A.

You might be interested in

Backpackers often use canisters of white gas to fuel a cooking stove's burner. If one canister contains 1.45 L of white gas, and

lorasvet [3.4K]

Answer: One canister contains 1.03 Kg of fuel,

Explanation:

The density is defined as the relation between the mass and the volume $p=\frac{m}{v}$

First of all you need to have the same units for volume and density, so:

For the volume,

$1.45 L * \frac{1.10*10^{-3}m^{3} }{1 L} = 1.45 *10^{-3} m^{3}$

For the density,

$p = 0.710\frac{g}{cm^{3} } *\frac{1cm^{3} }{1*10^{-6}m^{3} }*\frac{1 Kg}{1*10^{3}g } = 710\frac{Kg}{m^{3} }$

From the density equation we have $m=p*v$ , so:

$710\frac{Kg}{m^{3} } *1.45*10^{-3} m^{3} = 1.03Kg$

4 0

2 years ago

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

2 years ago

the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be

Nezavi [6.7K]

D:the electrons from being attracted to the grid instead of the anode

5 0

2 years ago

Read 2 more answers

>En cual de las siguientes situaciones la fuerza neta sobre el cuerpo es cero?

ExtremeBDS [4]

La respuesta es "Un avion que vuela al norte con rapidez constante y altitud constante".

Para que la fuerza neta sea 0, la aceleracion debe ser 0, para esto la velocidad debe ser constante.

Para que la velocidad sea constante el objecto debe estar moviendo con rapidez (magnitud de la velocidad) constante y sin cambiar direccion; ya que la velocidad es un vector asi es que depende en magnitud y direccion.

En las demas opciones la magnitud de la velocidad (rapidez) cambia y/o la direccion.

3 0

2 years ago

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete

Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(15)^2}{22.5}$

$a=10\ m/s^2$

We know that, $g=9.8\ m/s^2$

$a=\dfrac{10\times g}{9.8}$

$a=1.02\ g$

Hence, this is the required solution.

5 0

1 year ago