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rusak2 [61]
3 years ago
12

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

in x comes from slit width, ∆x=slit width?
Physics
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

0.53\times 10^{-30}kgms^{-1}

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

\Delta x\Delta p=\frac{h}{4\pi }

Given that the uncertainty in x is 0.1 mm.

Therefore,

\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}

Therefore, uncertainty in the transverse momentum of photon is 0.53\times 10^{-30}kgms^{-1}

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A boat travels at 30 mph for a huge, solid cliff that is about 3,000 meters away. When the horn on the boat makes a toot, you ca
AleksandrR [38]

Answer:The frequency of the echo is slightly decreased

Explanation:

Given

speed of boat =30\ mph

cliff is 3000\ m away

when boat is still , suppose t is the time taken by the echo to reach observer on the boat

But as soon as boat starts moving  the distance between cliff and boat decreasing and time for echo to reach observer also decreases

and we know time \propto \frac{1}{frequency}

therefore frequency of the echo slightly decreased.

5 0
3 years ago
If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
3 years ago
To balance the forces on the box what direction must you push?
ss7ja [257]
The correct answer would be left
3 0
3 years ago
What does the path of an object look like when it is in uniform motion?​
marysya [2.9K]

Answer:

The path of an object in uniform motion is a straight line.

4 0
2 years ago
Read 2 more answers
Nuclear decay occurs according to first-order kinetics. How long will it take for a sample of radon-218 to decay from 99 grams t
dedylja [7]

It will take 267 milliseconds for a sample of radon-218 to decay from 99 grams to 0. 50 grams.

We know that half life of a first order reaction is given by: t_{1/2} = 0.693/k

where k = rate of reaction

Given half life = 35 milliseconds

So from this we get k = 0.0198

Now we know that rate of first order reaction is given by: kt= 2.303 * log(R'/R)

where t= time

R'= initial amount = 99 g

R= final amount= 0.50 g

k= rate of reaction = 0.0198

Putting values of these in above equation we get t=267 milliseconds.

i.e. It will take 267 milliseconds for a sample of radon-218 to decay from 99 grams to 0. 50 grams.

To know more about radioactivity visit:

brainly.com/question/20039004

#SPJ4

4 0
2 years ago
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