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docker41 [41]
3 years ago
15

In each case, you should demonstrate how you worked out your answer, as well as giving the answer.

Physics
2 answers:
Ivanshal [37]3 years ago
7 0

Answer:

First frequency f = 153.846 Hz

Second period T = 1.625 ms

Step by stepplanation:

T = 6.5ms= 6.5*(10^-3)

But T = 1/f

f = 1/T

f = 1(6.5*(10^-3))

f = 153.846 Hz

Now the frequency f is increased by a factor of 3

= 3*153.846

= 461.538

New frequency = 461.538+153.846

= 615.384 Hz

New period T = 1/615.384

Period T = 1.625*10^-3

T= 1.625 ms

irga5000 [103]3 years ago
3 0

Answer:

Explanation:

(a)

Time period, T1 = 6.5 ms

= 6.5 × 10^-3 s

Frequency is the reciprocal of time period.

f1 = 1 / T1 = 1 / (6.5 × 10^-3) = 153.85 Hz

Now the frequency becomes 3 times

f2 = 3 × f1 = 3 × 153.85 = 461.55 Hz

The new period is

T2 = 1 / f2 = 1 / 461.55

= 0.0022 second = 2.22 ms

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denpristay [2]

The diameter of the wire is 2.8 * 10^-3 m.

<h3>What is the length?</h3>

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

A = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

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1 year ago
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sukhopar [10]

Answer:

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7 0
3 years ago
A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source
snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = 4.5 \times 10^{-5} C

Electric Field strength (E) = 2.0 \times 10^{4} N/C

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

U = q \times E \times d

So now insert the values to find the answer

U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m

On further solving

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