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storchak [24]
3 years ago
14

use the pendulum equation to calcuate the period of a 1.50 pendulum. Remeber that the vaule of "g" is 9.81 m/s² please help ​

Physics
1 answer:
kati45 [8]3 years ago
3 0

Answer:

L = 3.51

Explanation:

Pendulum equation is T = 2pi\sqrt{L/9.81}

T = 1.5 and we are solving for L

1.5=2\pi \sqrt{L/9.81}

square both sides to get 2.25 = 2\pi ( L/9.81)

multiply both sides by 9.81 then divide by 2 and 3.14 as a substitue for pi. The answer should be about 3.51 in length

L = 3.51

If this helps, mark me brainliest pls

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How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?
harkovskaia [24]

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

U = \mu B

where;

  • \mu is the dipole moment
  • B is the magnetic field

B = \frac{\mu_0 I}{2\pi r}

U = \mu\times  (\frac{\mu_0 I}{2\pi r} )

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

  • In step 1, to increase the potential energy, the iron will move towards the electromagnet.
  • In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.

Learn more about potential energy in magnetic field here: brainly.com/question/14383738

7 0
2 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
An electromagnetic wave has a frequency of 4.0 x 10^18 Hz. What is the wavelength of the wave?
LenaWriter [7]

Answer:

7.5 × 10^-11 m

Explanation:

Hope this helps !

3 0
3 years ago
Read 2 more answers
Explain<br> (0) what is meant by regeneration,<br> (ii) why an analogue signal cannot be regenerated
Kamila [148]

Explanation:

<h3>1.) Regeneration is the natural process of replacing or restoring damaged or missing cells, tissues, organs, and even entire body parts to full function in plants and animals.</h3>

2.) When noise is added to analogue signals, it usually sounds like background hiss. Such noise can not be removed so the original clean signal can not be re-created or re-generated.

6 0
3 years ago
You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b
Radda [10]

Answer:

\theta=34 \textdegree

Explanation:

From the question we are told that:

Mass m=55kg

Angle \theta =28.0

Coefficient of static friction \alpha =0.680

Generally, the equation for Newtons second Law is mathematically given by

For

\sum_y=0

N=mgcos \theta

for

\sum_x=0

F_{s}=mgsin\theta

Where

F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta

F_{s}=0.68*55*9.8*cos 28

F_{s}=323.62N

Therefore

\alpha mgcos \theta=mg sin \theta

\theta=tan^{-1}(0.68)

\theta=34 \textdegree

6 0
2 years ago
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