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3 years ago

HURRY PEASE what happens to current when there is a diode on the circuit and why?

Physics

1 answer:

kolbaska11 [484]3 years ago

6 0

Answer:

The current can't 'split down the parallel branch, because the diode is reverse biased so is blocking the flow of current. So basically it's acting as an open circuit. Also when the current flows it wouldn't reduce the currents amount flow through the resistor.

Explanation:

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The element niobium, which is a metal, is a superconductor (i.e., no electrical resistance) at temperatures below 9 K. However,

Andreyy89

Answer:

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4 0

3 years ago

Which one of the following types of electromagnetic radiation causes certain substances to fluoresce?

nalin [4]

The answer is D. <span>Uv causes more stuff to fouresce </span>

8 0

3 years ago

A water hose is used to fill a large cylindrical storage tank of

ludmilkaskok [199]

Answer:

maximum possible speed by solving above equation for 7D is

$v_{max} = \sqrt{\frac{49}{5}gD}$

minimum possible value of speed for solving x = 6D is given as

$v_{min} = \sqrt{9gD}$

Explanation:

Let the nozzle of the hose be at the origin. Then the nearest part of the rim of the tank is at (, ) = (6, 2) and the furthest part of the rim is at (, ) = (7, 2).

The trajectory of the water can be found as follows:

$x = (v_o cos45) t$

$y = (v_o sin45) t - \frac{1}{2}gt^2$

Now from above two equations we have

$y = x - \frac{gx^2}{v_o^2}$

now we know that height of the cylinder is 2D so we have

$x - \frac{gx^2}{v_o^2} = 2D$

by solving above equation we have

$x = \frac{v_o^2 \pm v_o^2\sqrt{1 - \frac{8gD}{v_o^2}}}{2g}$

now we know that maximum value of x is 7D

so the maximum possible speed by solving above equation for 7D is

$v_{max} = \sqrt{\frac{49}{5}gD}$

minimum possible value of speed for solving x = 6D is given as

$v_{min} = \sqrt{9gD}$

6 0

3 years ago

You can filter the particles out of

Art [367]

A is the answer do ur this question

5 0

3 years ago

The driver of a car traveling at 31.3 m/s applies the brakes and undergoes a constant deceleration of 1.6 m/s2.How many revoluti

lisov135 [29]

Answer:

$R=156.99\operatorname{Re}vs$

Explanation: The equations used are as follows:

$\begin{gathered} x(t)=x_o+v_ot+\frac{1}{2}at^2\Rightarrow(1) \\ v(t)=v_o+at\Rightarrow(2) \end{gathered}$

By using equation (2), the time needed for the car to come to rest is calculated as follows:

$\begin{gathered} v(t)=(31.3ms^{-1})_{}+(-1.6ms^{-2})t=0 \\ t=\frac{31.3ms^{-1}}{1.6ms^{-2}}=19.56s \\ t=19.563s \end{gathered}$

By using equation (1), The total distance traveled in that time would be as:

$\begin{gathered} x(19.563s)=_{}(31.3ms^{-1})\cdot(19.563s)+\frac{1}{2}(-1.6ms^{-2})\cdot(19.563s)^2\Rightarrow(1) \\ x(19.563s)=612.31-306.17=306.14m \\ \therefore\Rightarrow \\ x(19.563s)=306.14m \end{gathered}$

The revolutions taken by the tire before the car comes to rest would be:

$\begin{gathered} C=2\pi\cdot(0.31m)=1.95m \\ R=\frac{x(19.563s)}{C}=\frac{306.14m}{1.95m}=156.99\operatorname{Re}v \\ R=156.99\operatorname{Re}vs \end{gathered}$

3 0

2 years ago