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3 years ago

6

What picture of the electron is suggested by the quantum-mechanical model of the hydrogen atom? a. a raisin in pudding c. a plan

etary orbiting body b. a probability cloud d. a light quantum

1 answer:

Ivanshal [37]3 years ago

3 0

D. a light quantum this is the answer .

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If you connected 10 of these 12V (10 W) lamps in parallel, to the 12V source, how much current would the source have to supply

Flauer [41]

Answer:

$T=8.33A$

Explanation:

From the question we are told that:

Number of battery $n=10$

Voltage source $E=12V$

Lamp Power $P=10W$

Generally the equation for Resistance is mathematically given by

$R=\frac{V^2}{P}$

$R=\frac{12^2}{10}$

$R=14.4ohms$

Therefore

$R_{eq}=\frac{14.4}{10}$

$R_{eq}=1.44$

Generally the equation for Current is mathematically given by

$T=\ffrac{V}{Req}$

$T=\frac{12}{1.44}$

$T=8.33A$

6 0

2 years ago

If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the

Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0

2 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago

Which method is used to deliver heat in most central heating systems?

inessss [21]

The right answer is B. hope this helps you :)

4 0

3 years ago

Read 2 more answers

A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces.

Sunny_sXe [5.5K]

Answer: $240\ rad/s^2$

Explanation:

Given

Length of beam $l=2\ m$

mass of beam $m=5\ kg$

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

$\tau =F\times l=200\times 2=400\ N.m$

Also, the beam starts rotating about its center

So, the moment of inertia of the beam is

$I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2$

Torque is the product of moment of inertia and angular acceleration

$\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2$

7 0

3 years ago