Ask question

Ask question

All categories

3 years ago

10

You have a small piece of iron at 75 °C and place it into a large container of water at 25 °C. Which of these best explains what

will occur over time. A:The iron will cause the water to boil and turn to steam. B:The iron will take the heat from the water and get hotter. C:The water will cool significantly due to its colder temperature. D:Some heat from the iron will move to the water causing both to change temperatures.

2 answers:

jok3333 [9.3K]3 years ago

6 0

D:Some heat from the iron will move to the water causing both to change temperatures.

svp [43]3 years ago

3 0

D. Heat energy will be transferred within the system and if left long enough, there will be enough transferred energy to make both of them the same temperature.

You might be interested in

A ball is thrown straight upward and rises to a maximum

Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

$v_f^2 - v_i^2 = 2 a d$

here we know that

$v_f = 0$

also we have

$a = -9.81 m/s^2$

so we have

$0 - v_i^2 = 2(-9.81)(16)$

$v_i = 17.72 m/s$

Now we need to find the height where its speed becomes half of initial value

so we have

$v_f = 0.5 v_i$

now we have

$v_f^2 - v_i^2 = 2 a d$

$(0.5v_i)^2 - v_i^2 = 2(-9.81)h$

$-0.75v_i^2 = -19.62 h$

$0.75(17.72)^2 = 19.62 h$

$h = 12 m$

3 0

3 years ago

How do I do these? Please help

ICE Princess25 [194]

Explanation:

4a)the displacement is the distance moved in a direction but since no direction is given, the displacement is equal to the distance

b) the distance moved is 400m because that's the length of the track

4 0

3 years ago

What is most likely the color of the light whose second-order bright band forms an angle of 13.5° if the diffraction grating has

mash [69]

Answer:

the color is red.

8 0

3 years ago

Read 2 more answers

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers

In which of the following situations or places would you want to reduce the

sashaice [31]

Answer:

D

Explanation:

<em>The correct answer would be in the axle of the wheels while you ride your bicycle.</em>

Options A, B, and C requires that the forces of friction is increased in order to have more control.

However, option D requires that there is a minimal frictional force in the axle of the wheels of a bicycle while riding so that a little effort would be required to keep the bicycle moving.

<u>The lesser the friction, the lower the effort that would be needed to keep the bicycle moving and vice versa.</u>

4 0

3 years ago