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3 years ago

Plz help!!!

Physics

1 answer:

shtirl [24]3 years ago

3 0

Answer:

I think the answer to your question is true

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First law of equilibrium

valentina_108 [34]

Answer:

For an object to be an equilibrium it must be experiencing no acceleration.

Explanation:

Hope it helps.

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3 years ago

WILL GIVE BRAINLIEST !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

valina [46]

Gold is a soft metal with a number of properties..It is both malleable and ductille,Also it's a heavy metal ,It is soft,Yellow....etc. Hope that helps

6 0

3 years ago

Question 10 of 10

svetoff [14.1K]

Answer:

Explanation:

f=ma

200=2*m

divide both sides by 2

m=100

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3 years ago

Planets that are cold have only a small amount of gravity

Nonamiya [84]

There is no theoretical OR observational evidence for that statement.

4 0

4 years ago

Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold

arsen [322]

Answer:

Resistance of gold wire, $R=1977 \times 10^3 ohm$

Explanation:

In this question we have given

Density of gold, $d=19.3\times 10^3 \frac{kg}{m^3}$

resistivity of gold, $r=2.44\times 10^{-8} ohm.m$

Length of wire, $L= 2.05 km$

Temperature, $T= 20^oC$

We know that relation between volume and density is given as

$Density= \frac{mass}{Volume}$

Therefore, volume occupied by one gram gold is given as,

$V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3$ .........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

$V=\pi \times r^2 \times L$ ......(2)

Here,

$A= \pi \times r^2$ ...........(3)

here A is the cross sectional area of cylendrical gold wire

From equation 2 and 3

we got

$V=A \times L$ ...............(4)

on comparing equation 1 and equation 4, we got,

$A \times L=5.181\times 10^{-8} m^3$

$A=\frac{5.181\times 10^{-8} m^3}{2050 m}$

$A=2.53\times 10^{-11}m^2$

we know that resistance and resistivity are related by following formula,

$Resistance = resistivity\times \frac{L}{A}$ ................(5)

Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

$R=1977 \times 10^3 ohm$

Therefore resistance of gold wire, $R=1977 \times 10^3 ohm$

7 0

3 years ago