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Jet001 [13]
3 years ago
9

3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in

the middle of the square.The 4 electrons are held in place by some mechanism. The 4 electrons are released by the mechanism at the same time. They move and reach the corners of a square of side 0.8 meters, and keep on moving . Find the velocity of each electron at the corners of the square of side 0.8 meters.
Physics
1 answer:
Eduardwww [97]3 years ago
4 0

Explanation:

3

i believe that they are all going at 3.2 meters each, I did 4 times 0.8

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Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What i
OverLord2011 [107]

Answer: 2.89(10)^{-3} m

Explanation:

The <u>Heisenberg uncertainty principle</u> postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.  

In other words:  

It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.

Mathematically this principle is written as:

\Delta x \geq \frac{h}{4 \pi m \Delta V} (1)

Where:

\Delta x is the uncertainty in the position of the electron

h=6.626(10)^{-34}J.s is the Planck constant

m=9.11(10)^{-31}kg is the mass of the electron

\Delta V is the uncertainty in the velocity of the electron.

If we know the accuracy of the velocity is 0.001\% of the velocity of the electron V=2 km/s=2000 m/s, then \Delta V is:

\Delta V=2000 m/s(0.001\%)

\Delta V=2000 m/s(\frac{0.001}{100})

\Delta V=2(10)^{-2} m/s (2)

Now, the least possible uncertainty in position \Delta x_{min} is:

\Delta x_{min}=\frac{h}{4 \pi m \Delta V} (3)

\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)} (4)

Finally:

\Delta x_{min}=2.89(10)^{-3} m

5 0
3 years ago
According to the law of refraction, what is the relationship between the angle of incidence and the angle of reflection
Vedmedyk [2.9K]
The relation between the angle of incidence and the angle of refraction is known as Snell's Law. Since the light travels with different speed in different media, the direction of the ray of light will change when it crosses the boundary between the two media
7 0
3 years ago
Read 2 more answers
A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
Gnesinka [82]

Answer:

a).a_p=-2.39x10^{-12} rad/s^2

b).t=1016298.8 years

c).T_i=80.58x10^{-3}s

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

a_p=\frac{dw}{dt}

w=\frac{2\pi}{t}

a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}

dT=0.0808s

dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s

Replacing

a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2

b).

If the pulsar will continue to decelerate at this rate, it will  stop rotating at time:

t=\frac{w}{a_p}

w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s

t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s

t=1016298.8 years

c).

582 years ago to 2019

1437

T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s

5 0
4 years ago
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
n200080 [17]

Answer:

a) 4.9 s

b) 167.8 m

Explanation:

Hello!

To solve this question we need to make use of the equations of motion of both the motorcycle xm(t) and the car xc(t) at t=5

Let us consider the position of the motorcycle at t=5 as the origin, that is:

xm(t+5) = vt + (1/2)at^2

xc(t+5) = vt + 60 m

where v = 22.0m/s  and   a=5m/s^2

We are looking for the time t' when the position of the car and the motorcycle are the same:

xm(t'+5)=xc(t'+5)

vt' + (1/2)at'^2 = vt' +60m

t' = √(120 m /a) = 4.89898... s

Since we are considering the origin of the cooordinate system at the position when the motorcycle starts to accelerate, the distance travelled by the motorcycle until it catches the car is given by:

xm(t'+5)= vt' + (1/2)at'^2

xm(9.89898s) = (22 * 9.89898 + 2.5 * 9.89898^2)m

xm(9.89898s)= 167.777... m

3 0
3 years ago
When a particle vibrates and passes to the next vibration what happens?
kaheart [24]
Im not sure how detailed your answer is supposed to be but here it is: The particle that vibrates contains energy. That exact amount of energy is passed on to the next particle. This occurs because as you must know, energy is neither created nor destroyed. The same amount of energy contained in the vibration is simply passed on.
4 0
3 years ago
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