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2 years ago

After a wave passes through a medium, particles in the medium

Physics

1 answer:

sertanlavr [38]2 years ago

3 0

Answer:

If a particle is affected by a wave, then the particles are displaced. They move along the direction of the wave. Hence, After a wave passes through a medium, particles in the medium are moved along with the wave.

Explanation:

hope this helps you

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When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where i

Gnoma [55]

Answer:

The ball stops instantaneously at the topmost point of the motion.

Explanation:

Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.

The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.

The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.

4 0

3 years ago

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho

Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg) = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

$N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1$

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

= (500)(0.5)(0.1 × 10⁻⁶)

= 2.5 × 10⁻⁵A

C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

$P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}$

= 1250W

d) Final peak=

P= Ik/e

= $= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W$

P = 2.5 × 10⁷W

5 0

2 years ago

Read 2 more answers

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH = 1.06

Slav-nsk [51]

Answer:

$1.06085\times 10^{-10}\ m$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = 1 (ground state)

Angular momentum is given by

$L=mvr$

From Bohr's atomic model we have

$L=\dfrac{nh}{2\pi}$

$mvr=\dfrac{nh}{2\pi}\\\Rightarrow v=\dfrac{nh}{2\pi mr}$

The centripetal force will balance the electrostatic force

$\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow \dfrac{ke^2}{r}=mv^2\\\Rightarrow \dfrac{ke^2}{r}=m(\dfrac{nh}{2\pi mr})^2\\\Rightarrow r=\dfrac{n^2h^2}{4\pi^2mke^2}\\\Rightarrow r=\dfrac{1^2\times (6.626\times 10^{-34})^2}{4\pi^2 \times 9.11\times 10^{-31}\times 8.99\times 10^{9}\times (1.6\times 10^{-19})^2}\\\Rightarrow r=5.30426\times 10^{-11}\ m$

The diameter is $2\times 5.30426\times 10^{-11}=1.06085\times 10^{-10}\ m$

7 0

3 years ago

True or false. there would be no like in earth without the sun

alisha [4.7K]

hello!

True

have a goed day

6 0

2 years ago

A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is requ

Y_Kistochka [10]

Answer:

Power is 1061.67W

Explanation:

Power=force×distance/time

Power=65×9.8×15/9 assuming gravity=9.8m/s²

Power=3185/3=1061.67W

8 0

2 years ago