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kykrilka [37]
3 years ago
11

What is it that if you have, you want to share me, and if you share, you do not have?

Physics
2 answers:
Sever21 [200]3 years ago
8 0

Answer:

Secrets!!!

Explanation:

Colt1911 [192]3 years ago
4 0

Answer:

Secrets? Cuz if I share you the secrets then the secrets will no longer will be secrets.

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A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane
Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, \phi=8.6\times 10^{-3}\ T/m^2

The magnetic flux linked through the loop is :

\phi=B.A

\phi=BA\ cos\theta

Here, \theta=0

B=\dfrac{\phi}{A}

or

B=\dfrac{\phi}{\pi r^2}

B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0
3 years ago
Use Kepler's third law to find the planet's average distance (semimajor axis) from its star. (Hint: Because the mass of the star
konstantin123 [22]

Answer:

Explanation: The planet average distance = 42300km

Kepler's 3rd Law also known as the Harmonic Law states that;

for each planet orbitting the sun, its side real period squared divided by the cube of the semi-major axis of the orbit is a constant.

A planet, mass m, orbits the sun, mass M, in a circle of radius r and a period t. The net force on the planet is a centripetal force, and is caused the force of gravity between the sun and the planet.

Please find the attached file for the solution

3 0
3 years ago
The closely packed cones in the fovea of the eye have a diameter of about 2 μm. For the eye to discern two images on the fovea
Marina CMI [18]

Answer:

The distance between the object is l=0.0056\  cm

Explanation:

The free body diagram of this setup is on the first uploaded image

From the question

   The diameter of closely packed cones in the fovea of the eye is  =  2 \mu m

     The distance of separation by one cone(not excited ) is d = 4\mu m = 4*10^{-4}cm

     The distance between the two point-like object  is  l

     The diameter of the eye is D = 2 cm

     The distance of the two point-like object from the near point of the eye is A = 28 cm

 From the diagram we see that the light from the two point-like object form a triangle of similar base l and d  and height D and A

So for a triangle with similar base we have that

                \frac{l}{A} =\frac{d}{D}

                \frac{l}{28} = \frac{4*10^{-4}}{2}

making l the subject we have

            l = \frac{28 *4*10^{-4}}{2}

              l=0.0056\  cm

               

         

       

   

5 0
3 years ago
Read 2 more answers
NEED ANSWER PLEASE!!!!
olganol [36]

Answer:A, Concave

Explanation:

3 0
4 years ago
a rugby player tackles another player. in terms of kinetic energy and work done, why is it more difficult to tackle a heavier pl
matrenka [14]
The weight of the heavier player bakes it a lot harder to push him or do work because of the net force gravity<span />
6 0
3 years ago
Read 2 more answers
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