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Physics Middle School

When a photon collides with an electron and is deflected, the photon’s ____ decreases. 1. frequency 2. wavelength

2 months ago

Solution 1

Guest #2391800

2 months ago

Answer:

2. Wavelength

Explanation:

When a photon collides with an electron and is deflected, the photon’s wavelength decreases.

When photon gained energy by the collision of the electron. Its energy and frequency will be increased and its wavelength will be decreased.

Solution 2

Guest #2391801

2 months ago

Answer:

1. Frequency

Explanation:

When a photon collides with an electron, the effect of it is called Compton scattering, where the photons interact with the electron, losing momentum and therefore increase in wavelength and a decrease in frequency.

📚 Related Questions

Question

When the speed of electrons striking the anode is increased, the of the ______ of the emitted x-rays increases 1. frequency 2. density 3. wavelength

Solution 1

The speed of electrons striking the anode is increased, the density of the emitted x-rays increases.

Answer: Option 2

Explanation:

Whenever electrons are made to be incident on any anode, they will be interacting with the surface atoms of anode. So these interaction will lead to excitation of electrons of the surface atoms of the anode.

Thus when the excited electron will come back to ground state, it will emit x-rays. Thus, on increasing the speed of the striking electrons, it will gain the property of interacting with larger amount of surface atoms leading to increase in the emission of X-rays .

Thus, this increase in the emission of X-rays will lead to increase in the density of X-ray. So the speed of electrons striking the anode is increased, the density of the emitted x-rays increases.

Question

When the number of electrons striking the anode of an x-ray tube is increased, the ____ of the emitted X-Ray increases. 1. density 2. frequency

Solution 1

When the number of electrons striking the anode of an X-ray tube is increased, the density of the emitted x-ray increases

Question

What is the difference between analog and digital sound transmission?

Solution 1

In analog sound transmission, information is translated in electric pulses. In digital sound transmission, information is translated to binary codes of zero and one.

Explanation:

Electric pulses in analog sound transmission has a range of amplitude. Binary format of digital sound transmission has only two types of amplitudes.
The waves of analog transmission are sine waves while the waves in digital transmission are square waves.
Analog signals consume less bandwidth while digital signals consume more bandwidth

Question

9. How much work is done when a 15kg box is lifted to a height of 2 meters?

Solution 1

Answer: W = 294 J

Explanation: Solution:

Work is expressed as the product of force and the distance of the object.

W = Fd where F = mg

W= Fd

= mg d

= 15 kg ( 9.8 m/s²) ( 2m )

= 294 J

Question

Find the impulse of a 50. kg object under the following scenarios: a. The object accelerates to 7.5 m/s from rest. b. The object stops from a velocity of 12.0 m/s. C. The object change in velocity from 2.2 m/s to 6.3 d. The object hits the ground with 2.5 m/s and rebounds with the same speed

Solution 1

Answer:

a. $J = 375\, \rm N \cdot s$ .
b. $J = -600\; \rm N \cdot s$ .
c. $J = 205\; \rm N \cdot s$ .
d. $J = -250\; \rm N \cdot s$ .

Explanation:

When the velocity of an object changes, it would experience an impulse. If the mass of the object stays the same, and that the object moves along a line, the value of the impulse $J$ would be:

$J = m \cdot \Delta v$ , where

$m$ is the mass of the object, and
$\Delta v$ is the change in the velocity of the object.

On the other hand, the change in the object's velocity can be found with the equation:

$\Delta v = v_{\text{final}} - v_\text{initial}$ .

Note that if $m$ is in $\rm kg$ and $\Delta v$ is in $\rm m \cdot s^{-1}$ , the unit of $J$ would be $\rm N \cdot s$ .

a.

$v_\text{initial} = 0\; \rm m \cdot s^{-1}$ .

$v_\text{final} = 7.5\; \rm m \cdot s^{-1}$ .

$\Delta v = v_{\text{final}} - v_\text{initial} = 7.5\; \rm m \cdot s^{-1}$ .

$J = m \cdot \Delta v = 50 \times 7.5 = 375\; \rm N \cdot s$ .

b.

$v_\text{initial} = 12.0\; \rm m \cdot s^{-1}$ .

$v_\text{final} = 0\; \rm m \cdot s^{-1}$ .

$\Delta v = v_{\text{final}} - v_\text{initial} = -12.0\; \rm m \cdot s^{-1}$

$J = m \cdot \Delta v = 50 \times 12.0 = 600\; \rm N \cdot s$ .

c.

$v_\text{initial} = 2.2\; \rm m \cdot s^{-1}$ .

$v_\text{final} = 6.3\; \rm m \cdot s^{-1}$ .

$\Delta v = v_{\text{final}} - v_\text{initial} = 4.1\; \rm m \cdot s^{-1}$ .

$J = m \cdot \Delta v = 50 \times 4.1 = 201\; \rm N \cdot s$ .

d.

$v_\text{initial} = 2.5\; \rm m \cdot s^{-1}$ .

$v_\text{final} = -2.5\; \rm m \cdot s^{-1}$ .

Note that $v_\text{final}$ and $v_\text{initial}$ are of opposite signs. The reason is that the object's velocity has changed direction in this period.

$\Delta v = v_{\text{final}} - v_\text{initial} = -5.0\; \rm m \cdot s^{-1}$ .

$J = m \cdot \Delta v = 50 \times (-5.0) = -250\; \rm N \cdot s$ .

Question

Find the impulse of a 50. kg object under the following scenarios: a. The object accelerates to 7.5 m/s from rest. b. The object stops from a velocity of 12.0 m/s. c. The object change in velocity from 2.2 m/s to 6.3 m/s. d. The object hits the ground with 2.5 m/s and rebounds with the same speed

Solution 1

B. The object stops from a velocity of 12.0 m/s

Question

a bear has a mass of 500kg and 100,000 J of mechanical kinetic energy. what is the speed of the bear?

Solution 1

Answer: v = 20 m/s

Explanation: Solution:

Use the formula of Kinetic Energy and derive for v:

KE = 1/2 mv²

To find v:

v = √ 2 KE / m

= √ 2 ( 100000 J ) / 500 kg

= √ 400 m/s

= 20 m/s

Question

How does the thermometer measure your body temperature?

Solution 1

Answer:

when u put ur mouth on the thermometer it checks the heat of your mouth because the whole body has the same tempature so just by the te,p of your mouth it can tell ur body tempature

Explanation:

Question

A rocket launched into outer space travels 240000km during the 6.0 h after the launching. What is the average speed of the rocket in km/ h and m/s?

Solution 1

The average speed of rocket in km/h is 40000 km/h and in m/s is 11111 m/s.

Explanation:

As the rocket is launched into outer space, the distance is given as 240000 km during 6 h. So the average speed will be the sum of initial speed and final speed.

As initially, the speed of rocket will be zero when it is at rest. And when it reaches the final speed, it had covered 240000 km in 6 hr.

So final speed is given by,

$\text { final speed }=\frac{\text {distance}}{\text {time}}=\frac{240000}{6}=40000 \mathrm{km} / \mathrm{hr}$

Thus the average speed will be same as the final speed since the initial speed and time is zero.

Average speed =40000 km/h.

In order to represent it in m/s, we get average speed = $\frac{40000*1000}{3600} =11,111 m/s$

So the average speed of rocket in km/h is 40000 km/h and in m/s is 11111 m/s.

Question

Relationship between prism and lens

Solution 1

Answer:

In essence, optical lenses bend and focus light, known as refraction. Prism lenses, however, refract light a bit differently. ... Light passing through a prism will bend towards the base, while the image of the object viewed with the prism moves toward the peak.

Explanation:

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