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ankoles [38]
3 years ago
9

The proper mean lifetime of a muon is 2.2 × 10–6 s. A beam of muons is moving with speed 0.6c relative to an inertial observer.

Approximately how far will a muon in the beam travel, on average, before it decays?

Physics
1 answer:
Viefleur [7K]3 years ago
8 0
This might help you out

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A 40 cm wire with a radius of 3 cm is oriented along the y axis and carries a current of 2 A. What is the magnitude of the magne
ZanzabumX [31]

Answer:

a) B = 1.99 x 10⁻⁴ Tesla

b) B = 0.88 x 10⁻⁴ Tesla

Explanation:

According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:

B = μ₀ I L/4πr²

where,

μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹

I = current = 2 A

L = Length of wire = 40 cm = 0.4 m

a)

r = radius of magnetic field = 2 cm = 0.02 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²

<u>B = 1.99 x 10⁻⁴ Tesla</u>

<u></u>

b)

r = radius of magnetic field = 3 cm = 0.03 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²

<u>B = 0.88 x 10⁻⁴ Tesla</u>

7 0
3 years ago
Ionic compounds have high melting points because a lot of energy is needed to break the bonds between the ions.
velikii [3]
Its true the ionic compounds have a higher melting point
5 0
3 years ago
Read 2 more answers
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend
Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

5 0
4 years ago
Read 2 more answers
What are symptoms of Gomorrah
marishachu [46]
Consumed by Fire and Brimstone would be 1 symptom of Gomorrah lol
5 0
3 years ago
A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
Llana [10]

Answer:

a) C = 40.138\,pF, b) q = 16.056\,nC, c) U = 3.212\,\mu J

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

C = K\cdot \epsilon_{o}\cdot \frac{A}{d}

Where:

K - Dielectric constant.

\epsilon_{o} - Vaccum permitivity.

A - Plate area.

d - Distance between plates.

Hence, the capacitance of the system is:

C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)

C = 4.014\cdot 10^{-11}\,F

C = 40.138\,pF

b) The charge can be found by using the definition of capacitance:

q = C\cdot V_{batt}

q = (4.014\times 10^{-11}\,F)\cdot (400\,V)

q = 1.606\times 10^{-8}\,C

q = 16.056\,nC

c) The energy stored in the charged capacitor is:

U=\frac{1}{2}\cdot Q\cdot V_{batt}

U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)

U = 3.212\times 10^{-6}\,J

U = 3.212\,\mu J

3 0
3 years ago
Read 2 more answers
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