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Viktor [21]
4 years ago
14

Moving electrons are found to exhibit properties of

Physics
2 answers:
Korvikt [17]4 years ago
7 0

Answer:

(3) Both particles and waves

Explanation:

Burka [1]4 years ago
4 0

Answer:

(3) both particles and waves

Explanation:

When electrons are moving with certain speed then as per particle nature of electrons we know that when electrons collide to any surface or any electron then it will transfer its momentum to other surface.

Also during collision it will follow the rules of reflection i.e. the angle made with the normal before and after collision must be same.

All these characteristics will verify the particle nature of electrons.

Now as per de-broglie hypothesis the wavelength associated with moving electron is given by

\lambda = \frac{h}{mv}

so here its wave characteristics are also represented by the diffraction of moving electrons and the interference of electrons.

So here correct answer would be

(3) both particles and waves

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How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?
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If an ideal gas does not exist then why laws were stated?
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17. Which of the following is the closest weight in newtons of a 7.0-kilogram
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4 0
3 years ago
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0
3 years ago
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