All categories

3 years ago

Scientists use the micron as a unit of length for very

Physics

1 answer:

r-ruslan [8.4K]3 years ago

6 0

Answer:

1 micron = 1.00E-6 m is one way

1.00^-6 m is another but is not usually considered scientific notation, but

often convenient to use.

You might be interested in

3a and c someone please help I really don't understand

earnstyle [38]

F_x=Fcos(a)
F_y=Fsin(a)

1.F_x=166cos(55)
F_y=166sin(55) (55 because of the supplementary angle)

2.F_x= 575cos(-75) (complimentary angle/negative because its under x-axis)
F_y= 575sin(-75)

The values you get can be negative or positive depending on which side the vector is pointing at.

3 0

3 years ago

Two parallel wires carry equal currents in the opposite directions. Point A is midway between the wires, and B is an equal dista

zepelin [54]

Answer:

Complete Question:(missing part)

The current in the left wire having magnitude is = 210 A

The Current in the right wire having magnitude is = 10 A

Answer:

a. $B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )$

Ratio = B 1/B 2 = 210/10 = 21

$b.B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1}}{3}-I_{1}} )$

Ratio = B 1/B 2 = 210/3 X10 = 7

Explanation:

Ratio of the magnitude of the magnetic field at point A to that at point B =?

The current in the left wire having magnitude is = 210 A

The Current in the right wire having magnitude is = 10 A

distance between two wires = d

Given that both wires are carrying equal current but in opposite direction therefore

Magnetic field linked with the left wire at point A distance =d/2= according to biot savarts law

$B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{d}{2} }$

Magnetic field linked with the right wire at point B distance =d/2

$B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{d}{2} }$

Net magnetic field added

B net = B 1 + B 2

$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )$

Ratio = B 1/B 2 = 210/10 = 21

Magnetic field linked to left wire at point B(d+ d/2)

$B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{3d}{2} }$

Magnetic field linked to right wire at point B(d+ d/2)

$B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{3d}{2} }$

Net magnetic field subtracted

B net = B 1 - B 2

$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1} }{3}-1 } )$

Ratio = B 1/B 2 = 210/3 X10 = 7

4 0

3 years ago

A uniform magnetic field is perpendicular to the plane of a wire loop. If the loop accelerates in the direction of the field, wi

Nataly [62]

Answer:

Explanation:

If the coil is accelerated parallel to the magnetic field, it means also that the force on the coil acts parallel to the field. For current to be induced in a coil, a basic condition must be met which is that the directions of force (acceleration) and the magnetic field must be perpendicular to each other. This brings about current being induced in a direction perpendicular to bothering the directions of the force acting on the coil and the that of the magnetic field. This current would flow through the coil in either a clockwise or anticlockwise manner.

Since the coil is accelerated parallel to the magnetic field, no current is induced in it.

3 0

4 years ago

What is the nuclear binding energy in joules for an atom of helium–4? Assume the following: Mass defect = 5.0531 × 10-29 kilogra

Romashka [77]

The equation given, E = mc², can be directly used to determine the unknown amount of nuclear binding energy. Substitute the values given to the equation,
E = (5.0531 x 10^-29 kg) x (3 x 10^8 m/s)² = 4.548 x 10^-12 J
Thus, the nuclear binding energy is equal to 4.548 x 10^-12 J.

5 0

4 years ago

Read 2 more answers

An ice skater starts with a velocity of 2.25 m/s in a 50.0 degree direction. After 8.33s, she is moving 4.65 m/s in a 120 degree

Westkost [7]

Answer:

Explanation:

Lets decompose the initial velocity into its components:

Vi = 2.25 m/s

Vix = Vi x cos(50)

Viy = Vi x sin(50) = 2.25 x sin(50) = 1.724

Then decompose the final velocity:

Vf = 4.65

Vfx = Vf x cos(120)

Vfy = Vf x sin(120) = 4.65 x sin(120) = 4.027

After that we can use:

Vfy = Viy + ay*t

ay = (Vfy - Viy)/t

ay = (4.027 - 1.724) / 8.33

ay = 0.276

5 0

4 years ago