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3 years ago

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a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed

is equal to the minimum uncertainty possible in its speed

1 answer:

Elina [12.6K]3 years ago

7 0

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

$\Delta x \Delta p\geq \frac{h}{4\pi}$ (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

$\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}$

The minimum kinetic energy is calculated by using the following formula:

$k=\frac{(\Delta p)^2}{2m}$ (2)

m: mass of the proton = 1.67*10^{-27}kg

$k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J$

in eV you have:

$2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV$

The kinetic energy of the proton is 1.29eV

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B. transition metals for sure

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3 years ago

Ivan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Ivan drove h

lesya692 [45]

Answer:252 miles

Explanation:

Given

During his way to mountain it took 7 hr to drive

and during his return trip it took 4 hr to return

Let x be the distance between home and mountain

average speed for return is 27 miles per hour faster than his former trip

let v be the speed on his way to mountain thus v+27 is his return speed

thus $7=\frac{x}{v}$ ----1

for return trip

$4=\frac{x}{v+27}$ -----2

divide 1 & 2

$\frac{7}{4}=\frac{x\cdot (v+27)}{v\cdot x}$

$7v=4v+4\cdot 27$

$3v=4\cdot 27$

$v=36 mph$

thus $x=7\times 36=252\ miles$

7 0

4 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

3 years ago

A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?

andrey2020 [161]

Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19× $\frac{1}{2\pi} *60$ revolution per min

=728 rpm

Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

4 0

3 years ago

A beam of protons is directed in a straight line along the +z direction through a region of space in which there are crossed ele

IceJOKER [234]

Answer:

B = 7.6 T direction of + x

Explanation:

For the proton beam to continue in the same direction the electric and magnetic forces must be equal

$F_{m} - F_{e}$ = 0

$F_{m}$ = F_{e}

Fm = q E

The electric force is in the direction of the electric field because it is the charge of the positive proton, the electric force goes in the direction of –y, therefore, the magnetic force cancels this force must go in the direction of + y

The magnetic force is

F_{m} = q v x B = q v B sin θ

θ = 90

B = q E / q v

B = E / v

B = 800/105

B = 7.6 T

To find the direction of the magnetic field we use the right hand rule, the thumb goes in the direction of the proton velocity, the fingers extended in the direction of the magnetic field and the palm is the direction of force, for a positive charge.

Thumb goes in the direction of the + z axis

Palm in the direction of +y

Fingers point in the direction of + x

7 0

3 years ago