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anastassius [24]
3 years ago
15

According to Boyles’ law, PV = constant. If a graph is plotted with the pressure P against the volume V, the graph would be a(n)

: Question 2 options: A. straight line B. parabola C. hyperbola D. ellipse
Physics
2 answers:
Hunter-Best [27]3 years ago
4 0

Answer:

C. hyperbola

Explanation:

From Boyle's law:

PV = k, where k is a constant

Solving for P:

P = k / V

At first glance, this equation doesn't fit any of the options.  But when you graph it, you can see that it's actually a <em>rotated</em> hyperbola.

Ad libitum [116K]3 years ago
3 0

Answer:

C. hyperbola

Explanation:

According to Boyles’ law, PV = constant, if a graph is plotted with the pressure P against the volume V, the graph would be a(n) hyperbola.

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Answer:

The answer is radio waves

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A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not cons
igor_vitrenko [27]

Answer: (a) α = \frac{3I}{2.\pi.R^{3}}

(b) For r≤R: B(r) = μ_0.(\frac{I.r^{2}}{2.\pi.R^{3}})

For r≥R: B(r) = μ_0.(\frac{I}{2.\pi.r})

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

I_{c} = \int {J} \, dA

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr

I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr

I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}

\alpha = \frac{3I}{2.\pi.R^{3}}

For these circunstances, α = \frac{3I}{2.\pi.R^{3}}

(b) <u>Ampere's</u> <u>Law</u> to calculate magnetic field B is given by:

\int\ {B} \, dl = μ_0.I_{c}

(i) First, first find I_{c} for r ≤ R:

I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr

I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr

I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr

I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}

I_{c} = \frac{I.r^{3}}{R^{3}}

Calculating B(r), using Ampere's Law:

\int\ {B} \, dl = μ_0.I_{c}

B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} ).μ_0

B(r) = (\frac{Ir^{3}}{R^{3}2.\pi.r}).μ_0

B(r) = (\frac{Ir^{2}}{2.\pi.R^{3}} ).μ_0

For r ≤ R, magnetic field is B(r) = (\frac{Ir^{2}}{2.\pi.R^{3}} ).μ_0

(ii) For r ≥ R:

I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr

So, as calculated before:

I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}

I_{c} = I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = (\frac{I}{2.\pi.r} ).μ_0

For r ≥ R, magnetic field is; B(r) = (\frac{I}{2.\pi.r} ).μ_0.

3 0
3 years ago
For every 120 joules of energy input a car wastes 85 joules , find the useful energy output of the car ?
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Answer:

35 Joules

Explanation:

Applying

Input Energy(Q) = Useful energy output(U)+Wasted Energy(W)

Q = U+W.............................. Equation 1

Make U the subject of the equation

U = Q-W................... Equation 2

From the question,

Given: Q = 120 Joules, W = 85 Joules

Substitute these values into equation 2

U = 120-85

U = 35 Joules

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