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Andre45 [30]
3 years ago
13

What travels around the earth but stays in one spot?

Physics
1 answer:
mina [271]3 years ago
7 0
The sun?? It stays in one spot, but from our point of view, it travels around the earth...
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The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use
saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

M = \frac{Nl}{f_ef_0}

Where

N = Near point

l = distance between the object lens and eye lens

f_0= Focal length

f_e= Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

M = \frac{25*10}{3*5}

M = 16.67\approx 17\\

Therefore the correct answer is C.

3 0
3 years ago
Describe how an electromagnet can be switched on and off.​
Phoenix [80]

Answer:The magnetic field around an electromagnet is just the same as the one around a bar magnet. It can, however, be reversed by turning the battery around. Unlike bar magnets, which are permanent magnets, the magnetism of electromagnets can be turned on and off just by closing or opening the switch.

8 0
3 years ago
Which method will correctly determine whether the forces on an object are balanced or unbalanced?
MaRussiya [10]

For this case, the first thing you should do is define a reference system.

Once the system is defined, we must follow the following steps:

1) Do the sum of forces in a horizontal direction

2) Do the sum of forces in vertical direction

The forces will be balanced if for each direction the net force is equal to zero.

The forces will be unbalanced if for each direction the net force is nonzero.

Answer:

Add the forces in the horizontal and vertical directions separately.

6 0
3 years ago
Read 2 more answers
An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele
kirza4 [7]

Answer:

W = 462.5 keV

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

F = qE

F = (1.6 \times 10^{-19})(1.85 \times 10^6)

F = 2.96 \times 10^{-13} N

now the distance moved by the electron is given as

d = 0.25 m

so we have

W = F.d

W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)

W = 7.4 \times 10^{-14} J

now we have to convert it into keV units

so we have

1 keV = 1.6 \times 10^{-16} J

W = 462.5 keV

5 0
3 years ago
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vichka [17]
The answer to the problem b.
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