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bearhunter [10]

3 years ago

5

An object floats on water with 80% of its volume below the surface. the same object when placed in another liquid floats on that

liquid with 72% of its volume below the surface. determine the density of the unknown fluid.

1 answer:

GarryVolchara [31]3 years ago

8 0

0.9g/cm cubed, this is my full answer however i have to write at least 20 characters so this is filler :)

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Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul

Gnom [1K]

Answer:

$t=2.5\times 10^{-14}\ s$

Explanation:

We know that charge on electron

$q=1.6\times 10^{-19}\ C$

r= 2 nm

We know that force between two charge given

$F=K\dfrac{Q_1Q_2}{r^2}$

Now by putting the value

$F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}$

$F=5.67\times 10^{-11}\ N$

We know that mass of electron

The mass of electron

$m=9.1\times 10^{-31}\ kg$

F= m a

a= Acceleration of electron

a= F/m

$a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2$

$a=6.2\times 10^{19} m/s^2$

$S=ut+\dfrac{1}{2}at^2$

initial velocity given that zero ,u=0

$20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2$

$t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}$

$t=2.5\times 10^{-14}\ s$

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All of the members of a community belong to the same species tf

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That is false

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What do halo stars do differently from disk stars?

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2 years ago

Particles q1, 92, and q3 are in a straight line.

NNADVOKAT [17]

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

$\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}} = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N$

The force,by the charge q₂ on the q₃;

$\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}} = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09 \ N$

The net force is the sum of the two forces;

$\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N$

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

8 0

2 years ago