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3 years ago

When a river flows into an ocean, it slows down and deposits materials in its alluvial fan/delta

Physics

1 answer:

nikklg [1K]3 years ago

8 0

When a river flows into an ocean, it slows down and deposits materials in its delta

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Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

7 0

3 years ago

Which of these describes a real image?

Margaret [11]

Image from a far away object formed by a concave mirror

I have no idea but this is my best guess as a sophomore in college

8 0

3 years ago

Five groups of four vectors are shown below. All magnitudes of individual vectors are equal. Please rank the groups based on the

valkas [14]

With the addition of vectors we can find that the correct answer is:

C) Q> P > R = S > T

The addition of vectors must be done taking into account that they have modulus and direction. The analytical method is one of the easiest methods, the method to do it is:

Set a Cartesian coordinate system

Decompose vectors into their components in a Cartesian system

Perform the algebraic sums on each axis

Find the resultant vector using the Pythagoras' Theorem to find the modulus and trigonometry to find the direction.

In this exercise indicate that the modulus of all vectors is the same, suppose that the value of the modulus is A.

We fix a Cartesian coordinate system with the horizontal x axis and the vertical y axis, we can see that we do not need to perform any decomposition, so we perform the algebraic sums

Diagram P

x-axis

x = 2A

y-axis

y = 2A

The modulus of the resulting vector can be found with the Pythagorean Theorem

P = $\sqrt{x^2+y^2}$

P = $\sqrt{4A^2 +4A^2 }= \sqrt{8} \ A$

P = 2 √2 A

Diagram Q

x-axis

x = 3A

y-axis

y = A

Resulting

Q = $\sqrt{x^2+y^2}$

Q = $\sqrt{9A^2 + A^2 }$

Q = $\sqrt{10} \ A$

Diagram R

x- axis

x = 0

y-axis

y = 2 A

Resulting

R = $\sqrt{4A^2 + 0}$

R = $\sqrt{4} \ A$

Diagram S

x-axis

x = 2 A

y-axis

y = 0

Resulting

S = 2A

Diagram T

x- axis

x = 0

y-axis

y = 0

Resultant T = 0

We order the diagram from highest to lowest

Q> P> R = S> T

When reviewing the different answers, the correct one is:

C. Q> P> R = S> T

Learn more about adding vectors here:

brainly.com/question/14748235

5 0

2 years ago

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

He does whatever a spider can. Spider-Man, who has a mass of 76 [kg], is clinging onto an inclined wall forming an inclination a

Aliun [14]

Answer:570.54 N

Explanation:

Given

mass of man=76 kg

$\theta =50^{\circ}$

As man is standing over inclined building therefore

its weight has two components i.e. sin and cos component

Force perpendicular to inclined wall

$F=mgcos\theta =76\times 9.8\times \sin 50$

F=570.54 N

4 0

3 years ago