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3 years ago

.Express 3.0x10-4 -1.7x10-6 with proper significant figure.

Physics

1 answer:

Sav [38]3 years ago

5 0

Because of BOMDAS, you do the multiplication first.

(3.0x10) - 4 - (1.7x10) - 6

= 30 - 4 - 17 - 6

= 3

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If a wire lies withina magnetic field what must be true for the magnetic field to produce an electric current in the wire

BigorU [14]

Answer:

The magnetic field through the wire must be changing

Explanation:

According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ through it. This is stated mathematically as ε = dΦ/dt.

Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.

So, dΦ/dt = dABcosθ/dt

Since A and B are constant,

dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ

Since dθ/dt implies a change in the angle between A and B, since A is constant, it implies that B must be rotating.

So, <u>for an electric current (or voltage) to be produced in the wire, the magnetic field must be rotating or changing</u>.

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3 years ago

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Montano1993 [528]

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3 years ago

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A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial

Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

$qvB=m\frac{v^2}{r}$

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

For the ion of oxygen-16, we have:

$m_A=2.66\cdot 10^{-26}kg$

$q_A = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_A=2.90\cdot 10^6 m/s$

$B_A=1.30 T$

So the radius of its orbit is

$r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m$

For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

$B_B=1.30 T$

So the radius of its orbit is

$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

$d=2(r_B-r_A)=2(0.417-0.371)=0.092 m$

3 0

3 years ago

Is calculating the change of velocity the same as calculating acceleration?

koban [17]

Answer:

Yes! Thinking about it graphically a position vs time graph models meters per second in most cases, making every point on the line have the units m/s. If we want the find the slope we are finding the change between each point and those units would change to m/s/s or m/s^2 giving us the same units for acceleration. Simply put, slope of a velocity graph gives us acceleration.

Explanation:

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2 years ago

What are the basic si units for the speed of light?.

DanielleElmas [232]

C=meters/second or C=m/s

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2 years ago