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3 years ago

15

Can someone help me :)

1 answer:

klasskru [66]3 years ago

3 0

Answer:

-61 degrees F

Step-by-step explanation:

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Find total area of a regular pyramid with base of 2 and altitude of 3

Akimi4 [234]

You are not giving much information so the assumption is the base is 2 = meters and the height is 3 meters.<span>
Find total area of a regular pyramid with base of 2 and altitude of 3
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Surface Area = 16.64911 m²

3 0

3 years ago

Read 2 more answers

A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and

natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: $E=k\frac{q}{r^{2} }$ ; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: $p=\frac{q}{A}$ ; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: $S=4\pi r^{2}$ ; which gives us the surface of a sphere.

The inner surface: $Si=4\pi (0.06m)^{2} = 0.04 m^{2}$

The outer surface: $S=4\pi (0.08m)^{2}=0.08m^{2}$

Now we can calculate the charges,

Inner charge: $Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC$

Outer charge: $Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC$

Then, we are able to calculate both fields:

Inner field: $Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}$

Outer field: $Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}$

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0

3 years ago

Which sum or difference identity would you use to verify that cos (180° - q) = -cos q?

Phantasy [73]

Answer:

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Step-by-step explanation:

Given : $\cos (180^{\circ}-q)=-\cos q$

We have to write which identity we will use to prove the given statement.

Consider $\cos (180^{\circ}-q)=-\cos q$

Take left hand side of given expression $\cos (180^{\circ}-q)$

We know

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Comparing , we get, a= 180° and b = q

Substitute , we get,

$\cos (180^{\circ}-q)=\cos 180^{\circ} \cos (q)+\sin q \sin 180^{\circ}$

Also, we know $\sin 180^{\circ}=0$ and $\cos 180^{\circ}=-1$

Substitute, we get,

$\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0$

Simplify , we get,

$\cos (180^{\circ}-q)=-\cos (q)$

Hence, use difference identity to prove the given result.

7 0

3 years ago

Read 2 more answers

Half the difference between two numbers is 2. The sum of the

leva [86]

Answer:

as below

Step-by-step explanation:

1st the difference is 4 b/c half of 4 = 2

and

2*smaller +greater = 13

so check out small number that are 4 apart.. like 2 and 6

2(2)+6 =10

2(3) +7 = 13 aahh hah.. that's it

3 and 7 are the numbers being hunted :P

5 0

3 years ago

iogann1982 [59]

Answer:

THE ANSWER IS B

Step-by-step explanation:

NO EXPLANATION JUST TRUST ME

5 0

3 years ago