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sleet_krkn [62]
2 years ago
11

Show that a directed multigraph having no isolated vertices has an euler circuit if and only if the graph is weakly connected an

d the in-degree and out-degree of each vertex are equal
Physics
1 answer:
sergeinik [125]2 years ago
5 0
If we let
p as the directed multigraph that has no isolated vertices and has an Euler circuit
q as the graph that is weakly connected with the in-degree and out-degree of each vertex equal

The statement we have to prove is
p ←→q (for biconditional)
Since
p → q (assuming that p is strongly connected to q)
q ← p (since p is strongly connected to q)
Therefore, the bicondition is satisfied
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How does the sun transfer energy to Earth?
aleksley [76]

Answer:

By electromagnetic waves.

Explanation:

The sun transfers heat to earth via electromagnetic waves  in twomajor  ways:

Radiation- this is the transfer of energy by invisible electromagnetic ways.

Convection-The radiant sun energy warms the atmosphere and becomes heat energy. This transfer of heat through movement of fluids or usually air is called convection.

4 0
2 years ago
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4. A cylindrical tube has a length of 14.4cm and a radius of 1.5cm and is filled with a colorless gas. If the density of the gas
professor190 [17]

Answer:

Mass, m of gas is 0.2504 grams.

Explanation:

First, we need to solve for the volume of the cylindrical tube.

Volume of cylinder is given by the formula;

V = 2\Pi r^{2}h

Where, V represents volume.

π represents pie

r represents radius.

h represents height or length.

Given the following data;

Radius, r = 1.5cm

Length, h = 14.4cm

Density, d = 0.00123g/cm³

Substituting into the equation;

V = 2 * 3.142 * (1.5)^{2}14.4

V = 2 * 3.142 * 2.25 * 14.4

V = 203. 6016

Therefore, the volume of the cylindrical tube is 203. 6016cm³

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the formula;

Density = \frac{mass}{volume}

Mass = density  *  volume

Substituting into the equation, we have;

Mass = 0.00123 * 203. 6016

Mass = 0.2504g.

5 0
2 years ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Darya [45]

Answer:

y = -8.37 cm

Explanation:

As we know that the equation of SHM is given as

y = A cos(\omega t)

here we know that

\omega = \frac{2\pi}{T}

here we have

T = 0.66 s

now we have

\omega = \frac{2\pi}{0.66}

\omega = 3\pi

now we have

y = (8.8 cm) cos(3\pi t)

now at t = 2.3 s we have

y = (8.8 cm) cos(3\pi \times 2.3)

y = -8.37 cm

6 0
2 years ago
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If the magnitude of charges on these source charges is arranged in a descending order, which is the correct sequence?
bonufazy [111]
The source charges' magnitude is signified by the arrows pointing outward. The more arrows there are, the greater is its magnitude. This is because, each arrow represents an electrical force exerted by the source. When you add up all the arrows there is, the electrical force becomes even greater. The answer in descending order would be C > A > B > D.
6 0
2 years ago
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You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro
ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N

The spring obeys Hook's law:

F=k\Delta x

where k is the spring constant and \Delta x is the stretching of the spring. Since we know \Delta x=2.92 cm=0.0292 m, we can re-arrange the equation to find the spring constant:

k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance \Delta x is equal to the elastic potential energy stored by the spring, given by:

W=U=\frac{1}{2}k\Delta x^2

Substituting k=907.5 N/m and \Delta x=8.80 cm=0.088 m, we find the amount of work that must be done:

W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J

5 0
2 years ago
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