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3 years ago

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg

Physics

1 answer:

lukranit [14]3 years ago

3 0

Answer:

Mass of the steel cube = 7800 kg

Volume of the steel = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel = 7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

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

We are given with density and sides of the cube

then volume of the cube

= $(side)^3$

= $10^3$

= 1000 cubic centimetre

Now

$7.8 = \frac{mass}{1000}$

$mass = 7.8 \times 1000$

mass = 7800 kg

<u>(2)volume of steel:</u>

Given the mass = 8 kg

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

Substituting the values

$7.8 = \frac{8}{volume}$

$volume = \frac{8}{7.8}$

volume = 1.025 cubic centimetre

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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

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3 years ago

What force is required to accelerate a 385 kg couch at 0.2 m/s^2 ?

Juliette [100K]

Answer:

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2 years ago

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Nimfa-mama [501]

Answer:

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Explanation:

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F = 0.00000045356 N

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Answer:

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3 years ago

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The earth tilt about 23 degrees which leads to the formation of seasons on earth.

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This is the path as the earth rotates around its.

So we can conclude that the earth tilt about 23 degrees which leads to the formation of seasons on earth.

Learn more about season here: brainly.com/question/25870256

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2 years ago

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg​

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg