1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ugo [173]
3 years ago
8

I’m not sure how to solve this

Physics
1 answer:
spayn [35]3 years ago
6 0

Answer:

Option 10. 169.118 J/KgºC

Explanation:

From the question given above, the following data were obtained:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

You might be interested in
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
If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the
Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0
2 years ago
BLANK causes movement of an object, whereas BLANK is the work done on an object in a given amount of time.
Setler [38]
The first one is Force & the second one Power.
3 0
3 years ago
Read 2 more answers
Use Eq. cosϕ=R/Z to show that the average power delivered by the source in an L−R−C series circuit is given by Pav = I^2rmsR .
Evgen [1.6K]

Answer:

Explanation:

In a L C R circuit, the average power is given by

P_{av}=V_{rms}I_{rms}Cos\phi

As given in the question

CosФ = R / Z

And we know that

V_{rms}=I_{rms}\times Z

So

P_{av}=I_{rms}\times Z\times I_{rms}\times \frac{R}{Z}

P_{av}=I_{rms}^{2}\times R

6 0
3 years ago
A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t
Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

Ff = m*a   where Ff = μ * N    and a = \frac{V^2}{r}=\omega^2*r

N - m*g = 0   So, N = m*g.   Replacing everything on the original equation:

\mu*m*g = m*\omega^2*r   (eq2)

Solving for r:

r = \frac{\mu*g}{\omega^2}=1.41m

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0
3 years ago
Other questions:
  • in an isolated system bicyclist 1 and bicyclist 2 each with a mass of 80 kg, collide. bicyclist 1 was moving to the right at 8 m
    6·1 answer
  • HOW FAR CAN A PERSON RUN IN 10 MINUTES AT A SPEED OF 260M/MIN
    14·1 answer
  • What are the three fields E⃗ 1E→1, E⃗ 2E→2, and E⃗ 3E→3 created by the three charges? Write your answer for each as a vector in
    8·1 answer
  • Corrosion is the tendency of a metal to revert back to its natural state.
    8·1 answer
  • How much work did the movers do (horizontally) pushing a 170-kg crate 10.2 m across a rough floor without acceleration, if the e
    14·1 answer
  • What would the electrostatic force be for two 0.005C charges 50m apart?
    12·1 answer
  • Where would the barycenter of these two bodies be located given their masses?
    8·1 answer
  • Suppose that the speed of a ball moving in a horizontal circle is increasing at a steady rate. Is this increase in speed produce
    15·2 answers
  • explain why an inflated balloon is likely to burst if you leave it near a window with the sun streaming it?
    7·1 answer
  • Which ray diagram demonstrates the phenomenon of absorption?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!