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IRISSAK [1]
3 years ago
8

Describe an experiment that tests whether all metals are magnetic and identify the independent and dependent variable in your ex

periment.
Physics
1 answer:
scoray [572]3 years ago
7 0

Answer and Explanation:

NOTE: Magnetism means the magnetic property of a material that causes it to create a magnetic field, hence getting it attracted to a magnet.

EXPERIMENTAL PROCEDURE

1. Use a tape to attach a permanent magnet to the end of a ruler so that the magnet is facing away from the ruler. Don't cover the magnetic surface with the tape. ( Leave the magnet in its decorative casing.)

2. Place your metal objects in a row, and make predictions of which one of them will be attracted to the magnet and which will not.

3. Hold the magnet over each metals, and record which metals are attracted to the magnet. Go back over the

objects that were not affected by the magnet at least one more time to be sure you didn't miss any.

In this experiment, the independent variable is the magnetism of the magnet used. This is the independent variable because it remained unchanged and unaffected by the metals' magnetic properties all through the experiment.

While the dependent variable is the magnetism of the metals used. This is so because the magnetism of these metals varied and also because it is what is been measured in the experiment. Some were attracted to the magnet from very close range while others were attracted even at some centimeters away from the magnet which indicates that those metals have strong metallic properties.

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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
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Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

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E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

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