3. Does a wire with a current running through it

Physics

1 answer:

bagirrra123 [75]3 years ago

8 0

Answer:

as net charge of a current carrying conductor is zero so no electric field is associated with at.

And

as moving charges give rise to magnetic field so magnetic field is associated with it

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What is number 30,31 in this picture?

Maslowich

On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C

Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)

on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.

Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:

C, B, A and then D

the same idea as on Q30 applies to Q31 part b,

D,C,B then A

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3 years ago

How long will a radioactive isotope decay?

fiasKO [112]

Answer:

c - until it becomes a different, stable element

Explanation:

found it somewhere else than brainly you're welcome :)

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3 years ago

Read 2 more answers

"If E = 7.50V and r=0.45Ω, find the minimum value of the voltmeter resistance RV for which the voltmeter reading is within 1.0%

Virty [35]

Answer:

The minimum value is $R_V =44.552\ \Omega$

Explanation:

From the question we are given that

The voltage is $E = 7.5V$

The internal resistance is $r = 0.45$

The objective of this solution is to obtain the minimum value of the voltmeter resistance for which the voltmeter reading is within 1.0% of the emf of the battery

What is means is that we need to obtain voltmeter resistance such that

V = (100% -1%) of E

Where E is the e.m.f of the battery and V is the voltmeter reading

i.e V = 99% of E = 0.99 E = 7.425

Generally

E = V + ir

where ir is the internal potential difference of the voltmeter and

V is the voltmeter reading

Making i the subject of the formula above

$i = \frac{(E-V)}{r}$