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3 years ago

7

How does earths atmispher help it to sustain life

1 answer:

azamat3 years ago

7 0

First of all you spelled atmosphere wrong and 2 it sustains life by keeping the oxygen produced by the plants on earth for the humans and plants and animals to breath.

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mini stereo speaker cables are made of copper wire surrounded by robert electricity travel down the copper portion of the cable

Tom [10]

Here current is flowing through the copper wire so this shows that copper is good conductor of electricity.

It is having less resistance as it conducts the current easily

Now a rubber coating on it will protect us from electric shock

So this property shows that rubber is a bad conductor of electricity

It is having large electrical resistance due to which it will conduct no current

Rubber : - No transmittance of electricity

copper :- good transmittance of electricity

7 0

3 years ago

Just like energy is <br> matter is lost through an ecosystem

Novosadov [1.4K]

Yes energy is reduced

7 0

2 years ago

Read 2 more answers

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$

Try dealing with the powers of 10 first: On the right, we have

$\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}$

Meanwhile, the other values on the right reduce to

$\dfrac{6.67\times5.98}{3.8^2}\approx2.76$

Then taking units into account, we end up with the equation

$2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2$

Now we solve for $m_2$ :

$m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}$

$m_2=7.25\times10^{22}\,\mathrm{kg}$

or, if taking significant digits into account,

$m_2=7.3\times10^{22}\,\mathrm{kg}$

5 0

3 years ago

3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?

ZanzabumX [31]

Answer:

$E_s = 277.13V$

Explanation:

Given

$Load\ Voltage = 480V$

Required

Determine the voltage dropped in each stage.

The relation between the load voltage and the voltage dropped in each stage is

$E_l = E_s * \sqrt3$

Where

$E_l = 480$

So, we have:

$480 = E_s * \sqrt3$

Solve for $E_s$

$E_s = \frac{480}{\sqrt3}$

$E_s = \frac{480}{1.73205080757}$

$E_s = 277.128129211$

$E_s = 277.13V$

<em>Hence;</em>

<em>The voltage dropped at each phase is approximately 277.13V</em>

5 0

2 years ago

A concave mirror with a focal length of 32 cm produces an image whose distance from the mirror is one-third the object distance.

iVinArrow [24]

Answer:

part A ⇒ u = 1.28 m

part B ⇒v = 0.43 m

Explanation:

for u is the distance to the object from the mirror and v is the distance from the mirror to the image.

Part A:

the mirror equation is given by:

1/f = 1/v + 1/u

but we told that, v = 1/3u:

1/f = 3/v + 1/u = 4/u

1/f = 4/u

f = u/4

u = 4f

= 4×(32×10^-2)

= 1.28 m

Therefore the distance from the mirror to the object is 1.28 m.

part B:

v = 1/3×u = 1/3×(1.28) = 0.43 m

5 0

3 years ago