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kotegsom [21]
3 years ago
8

An electron in a TV picture tube is accelerated through a potential difference of 10 kV before it hits the screen. What is the k

inetic energy of the electron in electron volts?(1 eV = 1.6 ? 10 -19 J)
Physics
1 answer:
laiz [17]3 years ago
7 0

Answer:

10,000 eV

Explanation:

According to the law of conservation of energy, the kinetic energy gained by the electron is equal to its change in electric potential energy:

K=\Delta U=q \Delta V

where:

K is the kinetic energy of the electron

q=1 e is the magnitude of the charge of the electron

\Delta V is the potential difference through which the electron has been accelerated

For this electron in the TV, we have

\Delta V=10 kV=10000 V

Therefore, the kinetic energy of the electron in electronvolts is

K=(1 e)(10000 V)=10000 eV

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Ultrasonic images are obtained from the inside organs of our body. This process uses which property of sound wave?
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This question involves the concepts of echo, ultrasonic images, ultrasonic sound waves.

The process of ultrasonic images uses the "echo" property of the sound waves.

Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.

Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.

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2 years ago
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plan
pshichka [43]

Answer:

a)  P = 807.85 N,  b)  P = 392.15 N,  c)  P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

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          W_y = 1200 cos 30 = 1039.23 N

Y axis  

      N- W_y = 0

      N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

       P -Wₓ -fr = 0

       P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

      fr = μ N

      fr = 0.20 1039.23

      fr = 207.85 N

we substitute

       P = 600+ 207.85

       P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

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       fr = μ N

       fr = 0.20 1039.23

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we substitute

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c) as the movement is continuous, the friction coefficient is dynamic

         P - Wₓ + fr = 0

         P = Wₓ - fr

         fr = 0.15 1039.23

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         P = 600 - 155.88

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