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anzhelika [568]

3 years ago

9

What are common factors of 36 and 14

1 answer:

Step2247 [10]3 years ago

7 0

There's only one common factor for 36 and 14 and it is:

2

justification:
because the prime factorization of 36 is: 2 x 2 x 3 x 3
and the prime factorization of 14 is: 2 x 7

the only factor that they have in common is 2
therefore the common factor of 36 and 14 is 2

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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

Will mark brainliest

faust18 [17]

$x^3$ is strictly increasing on [0, 5], so

$\max\{x^3 \mid 0\le x\le5\} = 5^3 = 125$

and

$\min\{x^3 \mid 0 \le x\le5\} = 0^3 = 0$

so the integral is bounded between

$\displaystyle \boxed{0} \le \int_0^5x^3\,dx \le \boxed{125}$

8 0

2 years ago

I don’t understand this at all

Lesechka [4]

Step-by-step explanation:

I dont either tbh

6 0

3 years ago

I need help with this pls

krek1111 [17]

Answer:

<h3>Janet</h3>

Step-by-step explanation:

<h2>Because janet daily step count is 5000-9000</h2><h2>And brends step count is</h2><h3>7000-10000</h3>

<h3>Janat step count is 1000 greater than brenda.</h3>

<h3>Please mark me brainalist. </h3>

4 0

3 years ago

The graph translated horizontally two units right, vertically four units down and reflected across x-axis from its parent functi

Keith_Richards [23]

Parent Function: f(x) = x^2

Right 2 units: f(x) = (x^2 - 2)

Down 4 units: f(x) = (x^2 - 2) - 4

Reflection over x-axis: f(x) = - (x^2 -2) + 4

Answer: f(x) = - (x^2 -2) + 4

3 0

3 years ago