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Bess [88]
3 years ago
14

Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown

solution?
Physics
1 answer:
jonny [76]3 years ago
5 0

Answer:

Precipitation is the formation of a solid from a solution. It is necessary to centrifuge the precipitate to exert sufficient forces of gravity to bring the solid particles in the solution to come together and settle

Explanation:

When you centrifuge precipitate it enables the nucleation to form.

Centrifuging the precipitate helps in determining whether a certain element is present in a solution or not.

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A hurricanes energy comes from?
Shalnov [3]

Answer:

a or b

Explanation:

6 0
2 years ago
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Can anyone help me please​
ArbitrLikvidat [17]

Answer:

Gravity.

Rocket ships.

Ball.

Basketball.

Explanation:

Gravity has to do a lot with air. It puts the planets in there area.

Rocket Ship has to do a lot with air. If i'm right, they calculate the area, weather, about the air.

A ball gets throwed in the air, which gravity comes into place.

Basketball is also a similar example to a ball.

7 0
3 years ago
Help me quick!!! please!!
Rudik [331]

Answer:

39.240 W

Explanation:

Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is w= \vec F\cdot \vec{h} = mg h = 1000 \times 9.81\times 40 = 392.400 J

At this point we can simply apply the definition of power, that is P = \frac wt, to get the power of the engine is 39.240 W

4 0
2 years ago
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A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−
Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

8 0
3 years ago
) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar
ZanzabumX [31]
Momentum would be the same before and after the collision
 Before the collision:
 Momentum of the single cart: 1 * 0.50 = 0.50
 After the collision
 velocity = 0.25m / s
 1 * 0.25 + 1 * 0.25 =
 0.25 * (1 + 1) =
 0.25 * 2 =
 0.50
 Now new momentum will be 0.5
 answer
 the same before and after the collision
4 0
3 years ago
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