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Nuetrik [128]
2 years ago
9

If we treat the electron as a classical particle moving at a speed at the outer edge of the uncertainty range, how long would it

take to traverse the channel
Physics
1 answer:
Marina86 [1]2 years ago
8 0

The correct answer is 5.78*10^-5  

∆x∆v = 5.78*10^-5  

 ∆v = 1157.08 m/s

<h3>what is a transverse wave?</h3>

Transverse wave motion occurs when all points on a wave oscillate along pathways that are perpendicular to the wave's advance direction. Transverse waves include surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves.

A sine or cosine curve can be used to depict a simple transverse wave because the amplitude of any point on the curve—that is, its distance from the axis—is proportional to the sine (or cosine) of an angle. The graphic depicts sine curves of varying amplitudes. These curves depict how a standing transverse wave can appear at successive (1, 2, 3, 4, and 5) time intervals.

learn more about transverse wave refer:

brainly.com/question/13761336

#SPJ4

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Which of the following is true about the Sun?
bekas [8.4K]
A. The Sun is the only star in the Solar System. This is the only star we know of.
I hope this helps!!
6 0
2 years ago
When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is
padilas [110]

Answer:

P_{C} = 3.2\, atm

Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:

P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T

Bulb B (3 L, 4 atm) - Before opening:

P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T

Bulbs A & B (5 L) - After opening:

P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T

After some algebraic manipulation, a formula for final pressure is derived:

P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}

And final pressure is obtained:

P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}

P_{C} = 3.2\, atm

5 0
3 years ago
What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h
Citrus2011 [14]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
where
m is the mass of the substance
C_s is its specific heat capacity
\Delta T is the increase of temperature

The sample of silver of our problem has a mass of m=24.7 kg. Its specific heat capacity is C_s = 236 J/g^{\circ}C and the increase in temperature is
\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C

Therefore, the amount of heat needed is
Q=mC_s \Delta T=(24.7 kg)(236 J/g^{\circ}C)(16.0^{\circ}C)=9.32 \cdot 10^4 J
8 0
4 years ago
When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be ________
Nana76 [90]

Answer: Oblique impact

Explanation:

When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be oblique impact.

On the other hand, when the directions of motion of the two colliding particles are moving along a line of impact, then it's refered to as central impact.

7 0
3 years ago
I’m gym class you run 22m horizontal then climb a rope vertically for 4.8m. What is the direction angle of your total displaceme
Sliva [168]

The answer is: 12.30 degrees.

To solve this problem you must apply the proccedure shown below:

1. You run 22 meters horizontally and then you climb a rope vertically for 4.8 meters.

2. So. let's call the angle \alpha, therefore, you must find it using tan^{-1}, as following:

tan^{-1}(\alpha)=\frac{opposite}{adjacent} \\opposite=4.8\\adjacent=22

3. Substitute values and solve for the angle:

tan^{-1}(\alpha)=\frac{4.8}{22}\\\alpha=12.30degrees

5 0
3 years ago
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