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2 years ago

Read the following statement.

1 answer:

5 0

The statement "the knowledge produced by science builds on old ideas and is constantly changing" is absolutely agreeable.

We can see the above statement repeating itself in various situations and has led to the discovery of new laws which are now used in all areas of science. The best example of the above statement is The universal gas law, PV = nRT, which is the combination of two laws, Boyle's law, Charle's Law and Avogadro's Law.

According to Boyle's Law, gas volume grows as pressure lowers.

Charles' Law states that the gas expands in volume as its temperature rises.

As per Avogadro's Law, that as the gas concentration rises, so does its volume.

To know more about universal gas law, refer here

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Which device is based on the expansion of matter as temperature increases?

Nat2105 [25]

Thaattttttt would be mass vs volume im in science now. hope it helps

3 0

3 years ago

Read 2 more answers

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

3 years ago

(6.MS-ETS2-1(MA).) The electrons in __________ move about more freely than the electrons in insulators which is why this type of

Tresset [83]

Answer:

A) conductors

Explanation:

A conductor can be defined as any material or object that allows the free flow of current or electrons (charge) in one or more directions in an electrical circuit. Some examples of a conductor are metals, tungsten, copper, aluminum, iron, graphite, etc.

Basically, the main purpose of a conductor in physics is to provide a low-resistance path between electrical circuits or components. This low-resistance path is to ensure that the electrical components allows the free flow of electrons and thus, enabling charge transfer.

Hence, the electrons in conductors move about more freely than the electrons in insulators which is why this type of material can be used to create electric circuits because it would significantly provide a low-resistance path between the electric circuits.

8 0

3 years ago

What kind of weather is signaled by cumulus clouds

Novay_Z [31]

Cumulonimbus clouds are associated by severe weather. They are thunderstorm clouds that bring heavy rain, snow, hail, lightning and even tornadoes.

5 0

3 years ago

A football is thrown horizontally with an initial velocity of(16.6 {\rm m/s} ){\hat x}. Ignoring air resistance, the average acc

Ray Of Light [21]

Answer:

A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.

Explanation:

A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.

If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:

vₓ = 16.6 m/s i

In the vertical direction, the football, once released, is in free fall, starting from rest.

So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:

vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s

Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:

vy = -17.2 m/s j

So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:

v = 16.6 m/s i -17.2 m/s j

b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:

$v =\sqrt{(16.6m/s)^{2}+ (-17.2m/s)^{2}} = 23.9 m/s$

v = 23.9 m/s

c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:

tg θ = $\frac{-17.2}{16.6} =-1.036$

⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.

6 0

4 years ago