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Vadim26 [7]
2 years ago
11

A gas of helium atoms at 273 k is in a cubical container with 25.0 cm on a side. (a) what is the minimum uncertainty in momentum

components of helium atoms? (b) what is the minimum uncertainty in velocity components? (c) find the ratio of the uncertainties in (b) to the mean speed of an atom in each direction.
Physics
1 answer:
qwelly [4]2 years ago
6 0

wave function of a particle with mass m is given by ψ(x)={ Acosαx −

π

2α

≤x≤+

π

2α

0 otherwise , where α=1.00×1010/m.

(a) Find the normalization constant.

(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.

(c) Find the particle’s average position.

(d) Find its average momentum.

(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.

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A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direc
Alborosie

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is 6.21 \times 10^{-3} N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

           Q = 6.50 \times 10^{-6} C

           E = 1300 N/C in the +x direction

           B = 1.02 T in the +z direction

and,    F_{net} = 6.25 \times 10^{-3} N in the +x direction

Also,       F_{net} = F_{E} - F_{b}

                         = qE - qvB

Now, we will calculate the value of v as follows.

             v = (\frac{1}{B}) \times (E - \frac{F_{net}}{q})

                 = (\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})

                v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for F_{net}.

Using the right hand rule where:

your right thumb goes toward the F_{net}, then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that v_{x}, v_{y}, v_{z} = 0, -(458.507), 0.

8 0
2 years ago
A point charge Q = -400 nC and two unknown point charges, q1 and q2, are placed as shown. Point charge q1 is located 1.3 meters
nalin [4]
120 nC is the answer




Sorry if I’m wrong
6 0
2 years ago
All students except one are cheating on a test. The one student who is not cheating on the test is exhibiting abnormal behavior.
katen-ka-za [31]

Answer:

He may be experiencing abnormal behavior because his entire class is cheating on the test except This is considered abnormal because it is outside of the norm and is therefore abnormal

Explanation:

I put this on edenguity and got it right

7 0
2 years ago
Read 2 more answers
A force of 5000 n is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a young's modulus of 125 x 1
Blizzard [7]
Por definicion tenemos que
 (F/A) = E(∆/0)
 Sustituyendo los valores tenemos y despejando ∆:
 ∆ = (F/(πr2 × E))*0
 (5000×5)/(3.14×(34×10^−2)^2×(125×10^8))
 5.5×10^−6 m
 
4 0
2 years ago
Read 2 more answers
roblem 10: In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following p
miskamm [114]

Answer:

W= -2.5 (p₁*0.0012) joules

Explanation:

Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then

In adiabatic compression, work done by mixture during compression is

W= \int\limits^f_i {p} \, dV  where f= final volume and i =initial volume, p=pressure

p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ

W= \int\limits^f_i {K/V^} \, dV

W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)

W=1/1-γ (p₁Vf-p₀Vi)

W= 1/1-1.40 (p₁*6/5 -p₀*0)  

W= -2.5 (p₁*6/5*0.001)   changing liters to m³

W= -2.5 (p₁*0.0012) joules

3 0
2 years ago
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