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2 years ago

14

Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He

knows that the distance between the x-coordinates is 4 units. Which of the following fractions will let him find the x-coordinate for point K?
A. 2/3
B. 2/5
C. 3/2
D. 3/5

1 answer:

Rufina [12.5K]2 years ago

3 0

For the answer to the question above,
the distance from i to j is 5 parts
(2 parts from i to k and 3 parts from k to j)

The y distance from i to j is
10 - 2 = 8

Each part is 8/5 = 1.6
Therefore the distance between the 2 parts from i to k is 3.2

From the y coordinate of I which is 2 plus the 3.2 to point k
2 + 3.2 = 5.2

Answer y =5.2

Now just convert that to fraction and that will be the answer

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A cylinder containing an ideal gas has a volume of 2.6 m3 and a pressure of 1.5 × 105 Pa at a temperature of 300 K. The cylinder

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<h2>Answer: $13.5\times 10^{8}$ joules</h2>

Explanation:

From the first law of thermodynamics,

Δ $Q$ =Δ $U$ + $W$

Where $Q$ is the heat given to the gas,

$U$ is the internal energy of the gas,

$W$ is the workdone by the gas.

When pressure is constant,

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}$

When pressure is constant, $W=P$ Δ $V$

Where $P$ is pressure and $V$ is the volume of the gas.

Given $P=1.5\times 10^{5}Pa$

Δ $V=$ $7.8-2.6=5.2m^{3}$

So, $W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J$

Given that Δ $U=6\times 10^{5}$

So,Δ $Q=$ $6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J$

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3 years ago

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

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Explanation:

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Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

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$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

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