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sdas [7]
3 years ago
8

A 2.5m long steel piano wire has a diameter of 0.5cm how great is the tention in the wire if it stretches by 0.45cm when tighten

ed taking the young's modulus to be 2.0×10^11 N/m^2
Physics
1 answer:
UkoKoshka [18]3 years ago
7 0

Answer:                 Their u go i found it their was about 3 pages i did not no what pages u had to do.

Explanation:

Download pdf
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An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric
ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0
3 years ago
Create a Graph : Include numbers, axes, and order pairs
dlinn [17]
Answer:

a) Peak value = 20

b) Average value = 9.0

c) RMS value = 14.15

Explanation:

The peak-to-peak value = 40

v_{p-p}=40\begin{gathered} v_{p-p}=2v_p \\  \\ 40=2v_p \\  \\ v_p=\frac{40}{2} \\  \\ v_p=20 \end{gathered}

c) The rms value

\begin{gathered} v_{rms}=\frac{v_p}{\sqrt{2}} \\  \\ v_{rms}=\frac{20}{\sqrt{2}} \\  \\ v_{rms}=14.14 \end{gathered}

b) Average value over alternation of the sine wave

\begin{gathered} v_{avg}=0.637v_p \\  \\ v_{avg}=0.637\times14.14 \\  \\ v_{avg}=9.0 \end{gathered}

6 0
1 year ago
A charge of 50 µC is pushed by a force of 25 µN a distance of 15 m in an electric field. What is the electric potential differen
TiliK225 [7]
7.5 is the correct answer
8 0
3 years ago
Why us an element considered a pure substance
AVprozaik [17]

Explanation:

Hey there!!

An elements is called a pure substance because it is composed of only one kina of atoms in a fix proportion.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

6 0
3 years ago
Read 2 more answers
Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19
sergejj [24]

The distance between the charges is 13.86 X 10⁴m

<u>Explanation:</u>

Given:

Force, F = 1.2N

Charge, q₁ = 1.602 X 10⁻¹⁹ C

k = 8.987 X 10⁹ Nm²/C²

Distance, d = ?

According to Coulomb's law:

F = k\frac{q_1q_2}{r^2} \\\\

Substituting the value in the formula we get:

1.2 = 8.987X 10^9 X\frac{1.602 X 10^-^1^9 X 1.602 X 10^1^9}{r^2} \\\\1.2 = \frac{23.06 X 10^9}{r^2} \\\\r^2 = 19.22 X 10^9\\\\r = 13.86 X 10^4m

Therefore, the distance between the charges is 13.86 X 10⁴m

5 0
3 years ago
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