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3 years ago

What is urea? plz help me

Physics

2 answers:

Bezzdna [24]3 years ago

5 0

Answer:

It is a colorless crystalline compound which is the main nitrogenous breakdown product of protein metabolism in mammals and is excreted in urine.

Galina-37 [17]3 years ago

5 0

Answer:

Urea is a compound of nitrogen that is the chief solid substance dissolved in the urine of a mammal and is formed by the breaking down of protein.

(Hope this helps! Btw, I answered first. Brainliest please! :D)

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Air enters a compressor with a pressure of 14.7 lbf/in.2 , a temperature of 70F, and a volumetric flow rate of 40 ft3 /s. Air ex

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210 hp

Explanation:

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If two objects, like the eggs in the video, experience the same change in momentum but over time periods of

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Two parallel rods are each 0.69 m in length. They are attached at their centers to a spring that is initially neither stretched

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Answer:

Explanation:

Let the separation required be d .

Force between rod = 10⁻⁷ x 2 I₁ I₂ L / d

where I₁ and I₂ are current in them , d is distance of separation and L is length of wire .

Force between rod = 10⁻⁷ x 2 x 1200 x 1200 x .69 / d

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d = .1987 /3.9

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3 years ago

when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo

Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

$F = k\frac{|q_1|.|q_2|}{r^2}$

Where,

k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

$\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1$ ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

$q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}$

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

$\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6$ .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

$-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123$

$q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\$

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4 years ago

Two waves collide and the temporary combined waves that results is smaller that the original wave.

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Destructive interference is when two waves cancel each other out or when the crest of one wave passes through the trough of another wave.

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