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3 years ago

What observations characterize solar maximum?

Physics

1 answer:

MakcuM [25]3 years ago

3 0

That maximum is the biggest thing..

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What value do we use to describe acceleration due to gravity

andrey2020 [161]

Answer:

9.8 m/s2

Explanation:

In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.

Got it from the internet, hope it helps though ^^

7 0

3 years ago

HEY CAN YALL PLS HELP ME IN DIS!!!!!!!!!!!!!

Serga [27]

Answer:

Hey!!

Your answer is: 0.72

Explanation:

if 760=1 then...

550=x

=550÷760= 0.72 in two s.f

6 0

3 years ago

A 2kg object initially going 4m/s to the right is later going 8m/s. whats the change in velocity?

Nezavi [6.7K]

The change in velocity is +4 m/s to the right (or -4 m/s to the left).

The object's mass is irrelevant.

6 0

3 years ago

A unit of mass is ??

Effectus [21]

Hello,

Answer: kilogram

Further explaining: in science is used to measure weight of an object and used for accreditation.
Hope this helps!

3 0

3 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

3 years ago