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3 years ago

I'll make you brainliest! EaSyQuEsTiOn pleaze ASAP hurry! Make you brainliest Promise!!!

Physics

1 answer:

inysia [295]3 years ago

8 0

Answer:

It hit Earth like a really long time ago, so many rocks and soil should have been piled on top of the crater because of wind, rain, etc.

Also, it is really deep and really old so when it is old, it gets less visible to the naked eye.

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One cubic meter (1.0 m3

hammer [34]

Because the masses that you give are for blocks that are 1 cubic meter in volume, they also serve as the densities for the two metals that you are comparing.

mass = density*volume 

volume = (4/3)*pi*r^3 

volume of iron sphere = (4/3)*3.14*0.0201^3 = 3.40*10^-5 m^3 

mass of iron sphere = 7860* 3.40*10^-5 m^3 = 0.27 kg = mass of Aluminum Sphere 

Volume of Al Sphere = 0.27/2700 = 9.90*10^-5 m^3 

Radius = cube root (volume / (4/3) / pi) = 2.87 cm. 

I did this using the MS calculator, and I'm not 100% sure on the numerical answer, but the process is what you need to do to solve the problem. You should double check my answer.

hope this helped :)

8 0

3 years ago

What best illustrates that light behaves like particles

Ira Lisetskai [31]

The light particle conceived by Einstein is called a photon. The main point of his light quantum theory is the idea that light's energy is related to its oscillation frequency. Einstein believed light is a particle (photon) and the flow of photons is a wave.

3 0

3 years ago

Read 2 more answers

What is the definition for relative motion

xeze [42]

Relative motion is the calculation of the motion of an object with regard to some other moving object.

7 0

3 years ago

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

The electric field produced by a single-point charge is given by

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$