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3 years ago

Does the world actually end on April 21, 2021?

Physics

1 answer:

Setler [38]3 years ago

7 0

Answer:

No? Umm Hi Btw I Know You Might Now Give Me Brainliest But Can I Please Have Brainliest I Had Brainliest On My other accoun.t But The Next Day It Got hacke.d That's Why Im Asking I Kept Trying To Login To My Hack.ed Account Then A Thing That says Your account Got Haccked Shows Up Everytime But Now ITs NoT Saying My Account got hacke.d anymore It Just Wont Let Me in Now.

Explanation:

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Read this /https://www.carbonbrief.org/polar-bears-and-climate-change-what-does-the-science-say

DerKrebs [107]

Melting ice would damage this polar bears habitat meaning the polar bear may decrease

5 0

3 years ago

Examples of drawing packages

Marat540 [252]

Answer:

The answer are given above in attachment.

5 0

3 years ago

a sound pulse emitted underwater reflects off a school of fish and is detected at the same place 0.01 s later. how far away are

andrew-mc [135]

In fresh water sound waves travel at 1497m/s at 25 degrees, I'll assume that's the characteristics of the water.

If it's 0.01s then you need to divide the speed by 100 to get the, 14.97, however it gets there and back in that time so you need to halve it.
<u>7.485m</u>

5 0

3 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

A tennis ball is tossed upward with a speed of 3.0\,\dfrac{\text m}{\text s}3.0 s m 3, point, 0, start fraction, start text, m

MrRa [10]

Answer:

The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).

Explanation:

Hi there!

The equation for the velocity of an object thrown upward is the following:

v = v0 + g · t

Where:

v = velocity of the ball.

v0 = initial velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

t = time.

To find the velocity of the ball at t = 0.40 s, we have to replace "t" by 0.40 s in the equation:

v = v0 + g · t

v = 3.0 m/s - 9.8 m/s² · 0.40 s

v = -0.92 m/s

The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).

8 0

3 years ago

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