All categories

3 years ago

Can anyone help me with these questions? TIA! (Don’t actually answer please! :) )

Physics

1 answer:

nataly862011 [7]3 years ago

8 0

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7: </h2>

$\huge\text{Graphs:}$

The graph of

• The I-V for Ohmic Metal wire conductor at constant temperature always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1

• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2

<h2>_____________________________________ </h2><h2>Question 8: </h2>

$\Large\textbf{Diode:}$

A diode is a device that allows current to flow in only one direction.

$\Large\textbf{Forward and Reverse Biasing:}$

Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)

Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.

$\Large\textbf{Answer to the Question "Resistance"}$

The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.

<h2>_____________________________________ </h2><h2>Best Regards, </h2><h2>'Borz' </h2>

You might be interested in

Please help with both questions thank you!

Katen [24]

I think is:

(4)Inner core

(4)different Kingdome different species

hope this help :)

3 0

2 years ago

A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show

Lady bird [3.3K]

Answer:

change in height is 1.664 mm

Explanation:

Given data

drops = 3.00 mm

diameter = 5.00 cm = 0.05 mm

decrease = 350 cm^3

temperature = 95°C to 44.0°C

to find out

the decrease in millimeters in level

solution

we will calculate here change in volume so we can find how much level is decrease

change in volume = β v change in temp ...............1

here change in volume = area× height

so = $\pi$ /4 × d² h

so we can say change in volume = $\pi$ /4 × d² × change in height .......2

so from equation 1 and 2 we calculate change in height

( β(w) -β(g) )× v× change in temp = $\pi$ /4 × d² × change in height

change in height = 4 × ( β(w) -β(g) ) v× change in temp / $\pi$ /4 × d²

put all value here

change in height = 4 × ( 210 - 27 )(350 ) $10^{-6}$ × (95-44) / $\pi$ /4 × 0.05²

change in height is 1.664 mm

5 0

3 years ago

A block is attached to a horizontal spring. On top of this block rests another block. The two-block system slides back and forth

Minchanka [31]

Answer:

Angular frequency will increase

No change in the amplitude

Explanation:

At extreme end of the SHM the energy of the SHM is given by

$E = \frac{1}{2} (m_1 + m_2)\omega^2 A^2$

here we know that

$\omega^2 = \frac{k}{m_1 + m_2}$

now at the extreme end when one of the mass is removed from it

then in that case the angular frequency will change

$\omega'^2 = \frac{k}{m_1}$

So angular frequency will increase

but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block

so here no change in the amplitude

6 0

2 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0

3 years ago

Which of the following is NOT an example of a girl on a bike who is accelerating?

Serggg [28]

Answer:

the bike would not be moving

Explanation:

7 0

3 years ago