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Citrus2011 [14]
3 years ago
5

Look in the comments. I d a r e y o u

Physics
2 answers:
Y_Kistochka [10]3 years ago
8 0

Answer:

ok

Explanation:

Alika [10]3 years ago
3 0

Answer:

Where is the comments??

Explanation:

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What is the current when a typical static charge of 0.234~\mu\text{c}0.234 μc moves from your finger to a metal doorknob in 0.59
Black_prince [1.1K]
The current is defined as the quantity of charge Q that passes through a certain location in a time \Delta t:
I= \frac{Q}{\Delta t}
Using the data of the problem, we find:
I= \frac{Q}{\Delta t} = \frac{0.234 \mu C}{0.595 \mu s}=\frac{0.234\cdot 10^{-6} C}{0.595 \cdot 10^{-6} s}=0.39 A
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Walt and Mary are my Customers at this point.

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8 0
2 years ago
Which acceleration-time graph corresponds to the motion of the car if it moves toward the right, while slowing down at a steady
Bumek [7]

We define acceleration as the rate of change of the velocity

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If you have positive velocity and negative acceleration, your speed decreases.

Now you get the idea, we will see that the correct option is graph 1.

We know that the car moves towards the right (let's define this as "the car has positive velocity") and we also know that te car is slowing down constantly (thus the acceleration needs to be negative and constant).

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brainly.com/question/12550364

5 0
3 years ago
A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to
jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire  =3.7\ MA

b) Current at the center of the circular coil =2.48\times 10^{5}\ A

c) Current near the center of a solenoid 236.8\ A

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}

\mu=4\pi \times 10^{-7}\ \frac{henry}{m}

Plugging the values,

Conversion 1\times 10^6 A = 1\ MA,and 2cm=\frac{2}{100}=0.02\ m

I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA

⇒Magnetic Field at the center due to circular coil (at center) is given by,B=\frac{\mu\times I (N)}{2(a)}

So I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A

⇒Magnetic field due to the long solenoid,B=\mu\ nI=\mu (\frac{N}{l})I

Then I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A  

So the value of current are  3.7 MA,2.48\times 10^{5} A and 236.8\ A respectively.

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True because the picture below proves this....

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