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VikaD [51]
3 years ago
15

The observation location A is located at < 0.112, 0.247, 0 > m. The conventional current running through the wire is 7.5 a

mperes. In this exercise you will calculate the magnetic field at the observation location due only to segment 4 of the wire.
What is the direction of conventional current in this wire?
Physics
1 answer:
xenn [34]3 years ago
5 0

Answer:

Positive direction

Explanation:

We know that the direction of flow of current is always opposite the direction of the flow of electrons. Assuming that the electrons move in negative direction then the current will flow in the positive direction.

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larisa [96]

Answer:

wow i have know idea

Explanation:

5 0
3 years ago
How many significant digits are in the number 204.0920<br>​
monitta

Answer:

Seven

Explanation:

The rules for significant digits are:

  • Non-zero digits are always significant.
  • Zeros between significant digits are also significant.
  • Trailing zeros are significant only after a decimal point.

Here, the 2, 4, 9, and 2 are significant because they are non-zero digits.

The first two 0s are significant because they are between significant digits.

The last 0 is significant because it is a trailing zero after a decimal point.

Therefore, all seven digits are significant.

3 0
3 years ago
Read 2 more answers
A dust particle floats in front of a silent loudspeaker as shown in the figure. The loudspeaker is turned on and plays a constan
dybincka [34]

Answer:tbh idk

Explanation:

tbh idk

8 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.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
N LC circuit has an oscillation frequency of 105 Hz. If C = 0.1 F , then L must be about:
Umnica [9.8K]

Answer:

L = 22.97 H

Explanation:

Given that,

Capacitance, C=0.1\ \mu F=0.1\times 10^{-6}\ F

Oscillation frequency, f = 0.5 Hz

The frequency of an AC circuit is given by :

f=\dfrac{1}{2\pi \sqrt{LC} }

Where

L is impedance

f^2=\dfrac{1}{4\pi ^2LC}\\\\L=\dfrac{1}{4\pi ^2 f^2 C}\\\\\text{Putting all the values}\\\\L=\dfrac{1}{4\pi^2 \times (105)^2\times 0.1\times 10^{-6}}\\\\L=22.97\ H

So, the impedance of LC circuit 22.97 H.

7 0
2 years ago
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