The formation of the outer planets are affected by their distance from the sun with having them to maintain the lighter elements that they are composed of such as the hydrogen and helium, having them far away will also make their planet more cooler as the sun is distant from them.

6 0

3 years ago

What is the SI unit for speed?

Maksim231197 [3]

That is meters per second, same as velocity.

3 0

3 years ago

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A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago

It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f

Elena-2011 [213]

Answer:

B) the change in momentum.

Explanation:

The impulse is defined as the product between the force applied on an object (F) and the duration of the collision ( $\Delta t$ ):

$J=F \Delta t$ (1)

We can rewrite the force by using Newton's second law, as the product between mass (m) and acceleration (a):

$F=ma$

So, (1) becomes

$J=ma \Delta t$

Now we can also rewrite the acceleration as ratio between the change in velocity and change in time: $a=\frac{\Delta v}{\Delta t}$ . If we substitute into the previous equation, we find

$J=m\frac{\Delta v}{\Delta t}\Delta t=m\Delta v$

And the quantity $m\Delta v$ is equivalent to the change in momentum, $\Delta p$ .

6 0

3 years ago