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2 years ago

A ____ is one of the tiny dots of light that form a grid on your screen.

Physics

2 answers:

Degger [83]2 years ago

8 0

Hi there

The answer is pixel

Hope this helps

Marta_Voda [28]2 years ago

4 0

<span>A pixel is one of the tiny dots of light that form a grid on your screen. </span>

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object is placed 15cm in front of a plane mirror. If the mirror is moved further 5cm away from the object and the image is

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Answer:

The image will most likely be 20cm in front the mirror since the mirror was placed further 5cm.

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2 years ago

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3 years ago

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3 years ago

Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying

agasfer [191]

Answer:

The charge density in the system is $4.25*10^4C/m$

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

$r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C$

We need to asume here the number of free electrons in a copper conductor, at which is generally of $8.5 *10^{28}m^{-3}$

The equation to find the current is

$I = VenA$

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

$I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)$

$I= 22.11a$

Now to find the linear charge density, we know that

$I = \frac{Q}{t} \rightarrow Q = It$

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

$V_d = \frac{l}{t} \rightarrow l = V_d t$

The charge density is defined as

$\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}$

Replacing our values

$\lambda = \frac{22.11}{5.20*10{-4}}$

$\lambda= 4.25*10^4C/m$

Therefore the charge density in the system is $4.25*10^4C/m$

5 0

3 years ago

A cylinder contains 3.5 L of oxygen at 350 K and 2.7 atm . The gas is heated, causing a piston in the cylinder to move outward.

Effectus [21]

Answer:

The pressure is $P_2 = 1.84 \ a.t.m$

Explanation:

From the question we are told that

The first volume of is $v_1 = 3.5 \ L$

The first pressure is $P_1 = 2.7 \ a.t.m$

The first temperature is $T_1 = 350 \ K$

The new temperature is $T_2 = 620 \ K$

The new volume is $V_2 = 9.1 \ a.t.m$

Generally according to the combined gas law we have that

$\frac{P_1 V_1 }{T_1 } = \frac{P_2 V_2 }{T_2 }$

=> $P_2 = \frac{P_1 * V_1 * T_2 }{T_1 * V_2 }$

=> $P_2 = \frac{ 2.7 * 3.5 * 620 }{ 350 * 9.1 }$

=> $P_2 = 1.84 \ a.t.m$

4 0

3 years ago