Ask question

Ask question

All categories

3 years ago

15

Uranus has an orbital period of 84.07 years. In two or more complete sentences, explain how to calculate the average distance fr

om Uranus to the sun and then calculate it.

2 answers:

Veronika [31]3 years ago

6 0

One way to solve this is by doing the oppsite of how you find the orbitial period. The way you find the orbitial period is by raising the distance to the 3rd. Take the square root of the orbitial period which is 9.16 and multiplied it by 3 to get 27.506. <span>The Average Distance from Uranus to the Sun is 27.506.</span>

dem82 [27]3 years ago

3 0

Use kepler's 3rd law.T^2 = a^3

84.07^2 = a^3

7067.76 = a^3

19.19= a

<span>Uranus is 19.19 AU away from the sun.</span>

You might be interested in

If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish. given that the

Svetlanka [38]

here the charge density of metal plate is given as

$charge density = \sigma$

now the electric field is given Gauss law

$\int E. dA = \frac{q}{\epsilon_0}$

now here E = constant

so we will have

$E. \int dA = \frac{q}{\epsilon_0}$

Since total area on both sides of plate will be double and becomes 2A

$E. 2A = \frac{q}{\epsilon_0}$

$E = \frac{q/A}{2\epsilon_0}$

$E = \frac{\sigma}{2\epsilon_0}$

Now if we will find the electric field inside the metal plate

Then as we know that total charge inside the plate will always be zero

so we have

$E = 0$

7 0

3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$

This is the new external F force that we are looking for:

$F=Friction1(ground)+Friction2(AB)$

we know Friction2(AB) because we found that in the previous block so:

$F=Friction1(ground)+mg$ *μs

for the other friction we have to use the equation:

$Friction(ground)=N(ground)$ *μs

from y axis we have:

$N(ground)-w-Normal(AB)=0$

$N(ground)=w+Normal(AB)$

we found the value of Normal(AB) with the previous block so:

$N(ground)=mg+mg=2mg$

and:

$Friction(ground)=2mg$ *μs

$F=Friction(ground)+mg$ *μs

$F=2mg$ *μs+μs* $mg=3mg$ *μs

and since: μs* $mg=41.6N$

the new F force would be:

$F=3mg$ *μs $=41.6*3=124.8N$

4 0

3 years ago

A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi

algol13

Answer:

The time taken by the wave to travel along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire $\rho_{s}= 7.86\times10^{3}\ kg/m^{3}$

Density of copper wire $\rho_{c}=8.92\times10^{3}\ kg/m^3$

We need to calculate the linear density of steel wire

Using formula of linear density

$\mu_{s}=\rho_{s}A$

$\mu_{s}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{s}=5.00\times10^{-3}\ kg/m$

We need to calculate the linear density of copper wire

Using formula of linear density

$\mu_{c}=\rho_{s}A$

$\mu_{c}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{c}=5.67\times10^{-3}\ kg/m$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}$

$v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}$

$v_{s}=170.3\ m/s$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}$

$v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}$

$v_{c}=159.9\ m/s$

We need to calculate the time taken by the wave to travel along the combination of two wires

$t=t_{s}+t_{c}$

$t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}$

Put the value into the formula

$t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}$

$t=0.458\ sec$

$t=458\ ms$

Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.

4 0

4 years ago

a_______ is a region where the magnetic fields of a large number of atoms are lined up parallel to a magnets field.

patriot [66]

Answer:

Magnetic Domain

Explanation:

5 0

3 years ago

Read 2 more answers

The items in a mixture can be returned to their original form.<br> True<br> False

Grace [21]

I believe it’s false.

4 0

3 years ago

Read 2 more answers