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2 years ago

6

Which kind of air front forms when a warm air mass meets the area of a cooler air mass?

1 answer:

mihalych1998 [28]2 years ago

8 0

<span>Answer: "a cold front" .
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There is no relationship between the overtone series/harmonic series and the patterns established on a vibrating Chladni plate.

galben [10]

Answer:

There is a relationship between them.

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3 years ago

Which geologic feature would most likely be represented by contour lines drawn far apart from one another?

Anna [14]

It would most definitely not be a hill or a cliff. I believe it would be a plain.

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2 years ago

An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.

jekas [21]

Answer:

$\varphi_1= 796.25 N m^2/C$

$\varphi_2= 0 N m^2/C$

$\varphi_3=686.1 N m^2/C$

Explanation:

From the question we are told that

Electric field of intensity $E= 3.25 kN/C$

Rectangle parameter Width $W=0.350 m$ Length $L=0.700 m$

Angle to the normal $\angle=30.5 \textdegree$

Generally the equation for Electric flux at parallel to the yz plane $\varphi_1$ is mathematically given by

$\varphi_1=EA cos theta$

$\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0$

$\varphi_1= 796.25 N m^2/C$

Generally the equation for Electric flux at parallel to xy plane $\varphi_2$ is mathematically given by

$\varphi_2=EA cos theta$

$\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90$

$\varphi_2= 0 N m^2/C$

Generally the equation for Electric flux at angle 30 to x plane $\varphi_3$ is mathematically given by

$\varphi_3=EA cos theta$

$\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5$

$\varphi_3=686.072219 N m^2/C$

$\varphi_3=686.1 N m^2/C$

7 0

2 years ago

3. The main difference between speed and velocity involves

Zinaida [17]

The difference between them is that
velocity is the speed with a direction, whereas speed does not have a direction.

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3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

2 years ago