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Darina [25.2K]
4 years ago
11

The average lifetime of μ-mesons with a speed of 0.95c is measured to be 6 x 10^6 s. Find the average lifetime of μ-mesons in a

system in which they are at rest?
Physics
1 answer:
Mnenie [13.5K]4 years ago
7 0

Answer:

19.2*10^6 s

Explanation:

The equation for time dilation is:

t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}

Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:

t = \frac{6*10^6}{\sqrt{1-\frac{(0.95c)^2}{c^2}}} = 19.2*10^6 s

It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.

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Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example
iragen [17]

Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

8 0
3 years ago
Two students, Jenny and Cho, are investigating motion.
IgorLugansk [536]

Answer:

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Answer:

Option A

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D.   Satin Cloth

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avanturin [10]
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