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1 year ago

Which two statements about kinetic energy are true

Physics

1 answer:

gtnhenbr [62]1 year ago

8 0

Answer:

A, C

Explanation:

It can be found in objects like a swing. It creates kinetic energy when it is coming back or going away, while moving.

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__________ skids occur when you suddenly press too hard on the accelerator and the drive wheels lose traction.

Fynjy0 [20]

Answer:

Power Skids

Explanation:

Power Skid like the question suggested, is a skid that occurs when you suddenly press too hard on the accelerator and the drive wheels lose traction. Power Skid is one of the 4 types of Skids. The other 3 includes;

Cornering skid

Blowout skid and

Braking skid.

Power skid occurs mostly as a result of insanely high speed and or acceleration,when the accelerator is pressed on too suddenly causing the vehicle to be driven beyond its capabilities, and then to skid, ultimately.

4 0

2 years ago

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

2 years ago

In which layer of the sun is energy transfferred between atoms?

sasho [114]

Answer:

Core is the layer of the sun is energy transferred between atoms.

Explanation:

4 0

2 years ago

Savatey [412]

Answer:

Distance covered is equal to all the distance traveled.

So for example, if you go from A to B, and then from B to C, the total distance covered is AB + BC.

Displacement is equal to the difference between the final position and the initial position.

So if we go from A to B, the displacement is simply the line AB.

While if we go from A to B, and then from B to C, the displacement will be a segment that directly connects A and C, such that:

displacement = √( (AB)^2 + (BC)^2)

Now, if we want to find the points such that the magnitude of the distance covered is equal to the magnitude of the displacement, we need to look at the pairs that are directly connected by a straight line.

Those are:

A to B ( or B to A)

B to C (or C to B)

C to D (or D to C)

7 0

2 years ago

True or False: If the initial speed of an object is zero, then the initial position of the object must be the origin. Explain

lions [1.4K]

Answer:

False

Explanation:

It is not necessary, that when the initial speed of the object is zero then it must be at origin.

It may be the object starts motion from the left of origin.

4 0

2 years ago