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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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How is ultraviolet light and infrared light alike? How are they different?
swat32
Alike - both part of the light spectrum, cannot be seen with the naked eye

Different - IR is less energetic than UV, UV has a shorter wavelength than IR

IR - infrared
UV - ultraviolet
5 0
3 years ago
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
GaryK [48]

We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

y = \frac{m\lambda}{d}

Here,

m = Number of order bright fringe

\lambda = Wavelength

d = Distance between slits

Both distance are the same, then

y_1 = y_2

\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}

\frac{m_1\lambda_1}{m_2\lambda_2} =1

 Rearranging to find the second wavelength

m_1 \lambda_1 = m_2 \lambda _2

\lambda_2 = \frac{m_1\lambda_1}{m_2}

\lambda_2 = \frac{7(629)}{8}

\lambda_2 = 550.3nm

Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm

7 0
2 years ago
Hey can someone send me the answer to this
bagirrra123 [75]

Cars 'A' and 'C' look like they're moving at the same speed.  If their tracks are parallel, then they're also moving with the same velocity.

5 0
3 years ago
Boron (B) has an atomic number of 5 and an atomic mass of 11. Boron has _____.
Neko [114]

Answer:

the answer is 5 electrons

Explanation:

because its the same name as the amount of protons

6 0
3 years ago
Read 2 more answers
The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is
astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

E = \frac{T}{P*sin(\theta)}=  \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

6 0
3 years ago
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