Which instrument would play lower tones?

14) What is the velocity of a wave 5.35 x 10 cm long with a frequency of 16kHz

JulijaS [17]

Answer:

wavelength= 5.35×10cm=0.535m

f= 16000hz

v= wavelength × f= 0.535×16000=8560m/sec

6 0

3 years ago

A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a

wariber [46]

<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $\frac{V^{2} }{R}$

=> R = $\frac{V^{2} }{P}$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $\frac{V^{2} }{P}$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $\frac{12.0^{2} }{4}$

R₁ = $\frac{144}{4}$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $\frac{V^{2} }{P}$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $\frac{12.0^{2} }{4}$

R₂ = $\frac{144}{4}$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$\frac{1}{R_{X} }$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$\frac{1}{R_X}$ = $\frac{1}{36}$ + $\frac{1}{36}$

$\frac{1}{R_X}$ = $\frac{2}{36}$

Rₓ = $\frac{36}{2}$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $\frac{11.7}{18}$

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $\frac{0.3}{0.65}$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0

3 years ago

Read 2 more answers

Mo is on a baseball team and hears that a ball thrown at a 45 degree angle from the ground will travel the furthest distance. Ho

Galina-37 [17]

Answer:

Explanation:

Usually the angle between the y axis and x axis is 90° and we know that for furthest travel the degree angle must be 45° with the horizontal, Mo must release the ball about halfway between straight ahead and straight up

3 0

3 years ago

Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon

Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

$T^2=\frac{4\pi}{GM}r^3$

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

$T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min$

the comet takes around 16.63min

8 0

3 years ago

What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec

jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

$E = \sigma T^{4}$ (1)

Where $\sigma$ is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

$\lambda max = \frac{2.898x10^{-3} m. K}{T}$ (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

$E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}$ (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

$T = \frac{2.898x10^{-3} m. K}{\lambda max}$ (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), $\lambda max$ (450 nm) will be express in meters:

$450 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $4.5x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}$

$T = 6440 K$

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

$700 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $7x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}$

$T = 4140 K$

Comparison:

$\frac{6440 K}{4140 K} = 1.55$

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0

3 years ago