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Cloud [144]
4 years ago
6

Identify the magnetic north pole of Earth’s magnet?

Physics
2 answers:
choli [55]4 years ago
8 0
Yeah I wouldn’t have known lol
jarptica [38.1K]4 years ago
5 0
If am reading this question right the answer should be south pole
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Use the above picture to fill in the blanks for the following statement.
jenyasd209 [6]

Answer:

C + 2O ------ CO2

Explanation:

"One" element of Carbon combines with "Two" elements of Oxygen, to form "One" compound of Carbon dioxide.

I didn't really get what you meant but this is my guess of what you meant

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3 years ago
Which type of galaxy has arms that contain sites of active star formation and start close to a bulge in the center?
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normal spiral I think. I hope this helps

5 0
3 years ago
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Does the earth behave like a magnet how do you prove it
Arlecino [84]

Take any item that's magnetized.  With a sharpie or a drop of nail polish, make a small mark on one end of the item.  Then, hang it up by a thread, in such a way that its two ends are at the same height off the floor, and it's free to turn on the thread.  

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5 0
4 years ago
an object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second. the object strikes the ground 3.
Olegator [25]
You know that the formula for finding horizontal displacement is; 

s=uₓt 
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Therefore, the ball travelled 15 m. 

Hope I helped :) 

7 0
3 years ago
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A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle du
White raven [17]

Answer:

a. \phi _B=3.0528\times10^-^3T\ m^2\\ \\b. \phi_B=1.83299\times10^-^3T\ m^2\\\\c.\phi_B=0

Explanation:

#Consider a circular area of radius R=2.98cm in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

a. first, find the magnetic flux if the magnetic field has a magnitude of B=0.23T and points toward the +ve z-axis. The angle between the magnetic field and the area is \theta=0. Hence the magnetic flux:-

\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2

Hence flux magnitude in +z direction is 3.0528\times10^-^3T \ m^2

b. We now find the magnetic flux when the field has a magnitude of <em>B=0.230T</em> and points at an angle of \theta=53.1\textdegree from the +z direction.

Magnetic flux is calculated as:

\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2

Hence the flux at an angle of 53.1\textdegree is 1.83299\times 10^-^3T \ m^2

c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0

Hence the magnetic flux in the +y-direction is zero.

3 0
3 years ago
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