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3 years ago

What does the vibrating object do to the air particles around it ?

Physics

1 answer:

Westkost [7]3 years ago

7 0

Answer:

When an object vibrates, it causes movement in surrounding air molecules. These molecules bump into the molecules close to them, causing them to vibrate as well. This makes them bump into more nearby air molecules.

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The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane

umka21 [38]

Answer:

$\Delta p_x=7.75\times 10^{-25}\ kg-m/s$

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, $\Delta x=68\ pm=68\times 10^{-12}\ m$

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

$\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }$

Putting all the values, we get :

$\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s$

So, the momentum component of this electrons is greater than $7.75\times 10^{-25}\ kg-m/s$ .

6 0

4 years ago

Adding a catalyst to a reaction has an effect similar to

Irina18 [472]

I will say increasing temperature but you dont have that option on your list, so I would take B. Increasing Concentration.

3 0

3 years ago

Blank is the change in position of an object

mamaluj [8]

Displacement is your answer :)

6 0

3 years ago

A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hang

Romashka-Z-Leto [24]

Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/ $s^{2}$ . The correct answer is 4.90 m/ $s^{2}$

Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.

Let M = 1kg and m = 3 kg

Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.

When the mass is released, the cart accelerates to the right can be calculated from Newton' second law of motion. That is,

M( g + a) = m(g - a)

1(9.8 + a) = 3( 9.8 - a)

9.8 +a = 29.4 - 3a

collect the like terms

4a = 19.6

a = 19.6/4

a = 4.9 m/ $s^{2}$

Therefore, the cart accelerates to the right with a magnitude of 4.9 m/ $s^{2}$ . The correct answer is 4.90 m/ $s^{2}$

Learn more about dynamics here: brainly.com/question/24994188

5 0

3 years ago

What is the frequency of a clock waveform whose period is 750 microseconds?

Allushta [10]

Use this formula to find your answer...

Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms

frequency (f)=1/( Time period).

Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.

Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.

6 0

3 years ago