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3 years ago

If san andreas fault slips 2.5 cm each year how far would it separate in 100 years

Physics

1 answer:

nadya68 [22]3 years ago

7 0

Answer:

250cm or 2.5m

Explanation:

using direct variation

the length is = l

the time (year) = t

let k be the constant

so he have:

l=kt

2.5=k× 1

k= 2.5

Now in the next statement the 100 is given but not the length so

l=kt

~ l=k× 100 years

But k = 2.5

so :

l= 2.5×100

l= 250 cm

Hence it will be 250cm or 2.5m at which they will get separated

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Answer:

9.73 x 10⁻¹⁰ m

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Uncertainty in position x uncertainty in momentum ≥ h / 4π

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= 2 x 10⁻⁶ x 3 / 100

= 6 x 10⁻⁸

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2 years ago

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu

galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

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Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

$F=I \times B \times L \times sin(\theta)$

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(b)

$\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N$

(c)

$\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N$

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