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3 years ago

6

house current is 120 volts. if a light bulb runs a current of 0.5 amps, what is the resistance of the bulb

1 answer:

Sauron [17]3 years ago

6 0

Ohm's Law tells the relationship between voltage, current, and resistance.
It can be written in three different ways, depending on which ones you know,
and which one you want to find.

Here's the one we need:

   Resistance = (voltage) divided by (current)

   = (120 V) / (0.5 Amp)

   =    240 ohms .

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An emu moving with constant acceleration covers the distance between two points that are 92 m

nasty-shy [4]

Answer:

a) $V_{o}=14.30 m/s$

b) $a=-0.046 m/s^{2}$

Explanation:

The complete question is written below:

An emu moving with constant acceleration covers the distance between two points that are 92 m apart in 6.5s. Its speed as it passes the second point is 14 m/s. What are (a) its speed at the first point and (b) its acceleration?

Since we are talking about constant acceleration, we can use the following equations:

$d=x-x_{o}=(\frac{V_{o}-V}{2})t$ (1)

$V=V_{o}+at$ (2)

Where:

$d=92 m$ is the distance between the two points

$V_{o}$ is the velocity of the emu at the first point

$V=14 m/s$ is the velocity of the emu at the second point

$t=6.5 s$ is the time it takes to the emu to cover the distance $d$

$a$ is the emu's constant acceleration

Knowing this, let's begin with the answers:

<h2>a) Speed at the first point</h2>

In this situation wi will use equation (1):

$d=(\frac{V_{o}-V}{2})t$ (1)

Finding $V_{o}$ :

$V_{o}=\frac{2d}{t}-V$ (3)

$V_{o}=\frac{2(92 m)}{6.5 s}-14 m/s$ (4)

$V_{o}=14.30 m/s$ (5)

<h2>b) Emu's acceleration</h2>

Now we will substitute (5) in equation (2):

$14 m/s=14.30 m/s+a(6.5 s)$ (6)

Finding $a$ :

$a=-0.046 m/s^{2}$ (7) This means the emu is decreasing its speed at a constant rate.

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Answer:

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Explanation:

The function which is inverse to exponentiation is called logarithm.

It is of the form

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The 10 here is the base of the logarithm. The logarithm with base 10 is called common logarithm

$log_{10}1000000=y\\\Rightarrow 10^y=1000000\\\Rightarrow 10^6=1000000\\\Rightarrow y=6$

So, log(1,000,000) = 6

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