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2 years ago

9

In physics, for a given object, work divided by distance will equal

1 answer:

Genrish500 [490]2 years ago

5 0

It is force because work/distance= force

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What are gamma rays? How are they used?

Karo-lina-s [1.5K]

I know it, X-rays<span>, </span>gamma rays<span> and beta particles are all </span>used<span> in medicine to treat internal organs. X-</span>rays<span> are produced by firing electrons at a metal target and </span>gamma rays<span> are emitted by the nucleus of radioactive atoms. </span>Gamma rays<span> are </span>used<span> to kill cancer cells, to sterilise medical equipment and in radioactive tracers.</span>

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2 years ago

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Through the process of blank

maksim [4K]

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Subduction, Trench, Mantle

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2 years ago

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What is the value of x in the equation below 1+2e^x+1=9

GuDViN [60]

<h2>Answer: 1.252</h2>

Explanation:

We are given this equation and we need to find the value of $x$ :

$1+2e^x+1=9$ (1)

Firstly, we have to clear $x$ :

$2e^x=9-1-1$

$2e^x=7$

$e^x=\frac{7}{2}$ (2)

Applying<u> Natural Logarithm</u> on both sides of the equation (2):

$ln(e^x)=ln(\frac{7}{2})$ (3)

$xln(e)=ln(\frac{7}{2})$ (4)

According to the Natural Logarithm rules $xln(e)=x$ , so (4) can be written as:

$x=ln(\frac{7}{2})$ (5)

Finally:

$x=1.252$

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3 years ago

The cunninghams are moving across the country. Mr.cunningham leaves 2.5 hours before mrs.cunningham. If he averages 40 mph and s

Eva8 [605]

When she starts out, he is (40x2.5)= 100 miles ahead of her.

She gains (65-40)= 25 miles on him every hour.

It takes her (100/25)= 4 hours to catch up to him.

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3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

2 years ago