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Verdich [7]
3 years ago
7

PLEASE HELP WILL GIVE 10 POINTS!!!!

Physics
2 answers:
frozen [14]3 years ago
7 0
If the data don't support your hypothesis, you don't just go and look for some other experiment that will support it. You dig deeper into this experiment, find out why, and maybe tweak the hypothesis. Eventually you wind up with a hypothesis that IS supported by experiment. Then you've got yourself a theory ! The choice is C.
jolli1 [7]3 years ago
7 0

Answer:

c

Explanation:

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a=vf-vi/t is the equation for calculating the acceleration of an object. write out the relationship shown in the equation using
nevsk [136]

Answer:

Acceleration is defined as the ratio between the change in velocity of an object and the time it takes for the object to change the velocity.

In formula, this is written as:

a=\frac{v_f -v_i}{t}

where

vf is the final velocity

vi is the initial velocity

t is the time it takes for the object to accelerate from u to v

Note that acceleration is a vector, so it can also be caused by a change in the direction of the velocity, not only by a change in its magnitude.

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You push a desk with 230 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction
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3 years ago
A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi
svetlana [45]

Answer:

E=0 at r < R;

E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}} at 2R > r > R;

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}} at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon} (integral over a closed surface)

where,

E = Electric field

Q_{enclosed} = charged enclosed within the closed surface

\epsilon = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

Q_{enclosed} = 0 and hence E = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

Q_{enclosed} = Q,

therefore,

E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to 4\pi r^{2})

or, E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}

at r >= 2R

Q_{enclosed} = 2Q

Hence, by similar calculations, we get,

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}

4 0
4 years ago
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