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3 years ago

Is V directly proportional to I (From V=IR)??Or is V inversely proportional to I (From P=IV) ???

1 answer:

3 0

First of all, I is proportional V according to the Ohm's Law. R is merely a constant you need to obtain an equation. However, it is true that R changes with temperature and pressure, therefore Ohm's Law is only applicable in an invariable environment. Also this constant R is different for different materials.

So, do not get confused.

Ohm's law is not a universal law, please remember that as well. Some materials do not follow it and we call them non-ohmic conductors. I hope I helped! ^-^

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A scientist testing to see how a cat “always” lands on their feet drops a cat upside down out of a third story window at a heigh

makkiz [27]

Answer:

Cat has 2.02 seconds to right itself.

Explanation:

Initial height of cat from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So, cat has 2.02 seconds to right itself.

7 0

3 years ago

What is the current in a 160V circuit if the resistance is 2Ω? V= I= R=

Alex787 [66]

Explanation:

v = IR

v= 160 R = 2

160 = 2I

2 2

I = 80A

4 0

3 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$