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MissTica
3 years ago
15

Which law describes the speed at which objects travel at different points in their orbits?

Physics
1 answer:
natali 33 [55]3 years ago
4 0
Kepler's 2nd law describes the speed at which objects travel at different points in their orbit
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Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown
jonny [76]

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.

5 0
3 years ago
PLS HELP !!
Ira Lisetskai [31]
A because it’s basic kinetic example
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Choose all the answers that apply. Models are used to observe phenomena which _____. happen too slowly} happen too quickly} are
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your answer is phenomena which are too large


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3 years ago
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The magnitude of vector A is 10. Which of the following could be the components of A?
kotykmax [81]

Answer: B. Ax = 6, Ay=8

Explanation:

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3 years ago
If an object is thrown in an upward direction from the top of a building 1.6 x 102 ft. high at an initial velocity of 21.82 mi/h
stiv31 [10]
In this case, the object is thrown upwards from the building. Therefore, it first achieves some height before its starts dropping.

Now, when going upwards
v^2 = u^2 - 2gs

Where,
v = final velocity
u = initial velocity
g = gravitational acceleration
s = height achieved from the top of he bulding

Using the values given;
v = 0 (comes into rest before it starts dropping)
u = 21.82 mi/h = 32 ft/s
g = 9.81 m/s^2 = 32.174 ft/s^2

Then,
0^2 = 32^2 - 2*32.174*s
32^2 = 2*32.174*s
s = (32^2)/(2*32.174) = 15.91 ft

After achieving that height, it starts to drop from rest to maximum velocity when it hits the ground.
Applying the same formula;
v^2 = u^2 + 2gs

Where;
v = velocity when it hits the ground
u = initial velocity, 0 ft/s as it starts from rest
s = 15.91+1.6*10^2 = 15.91+160 = 175.91 ft

Therefore,
v^2 = 0^2 + 2*32.174*175.91
v^2 = 11319.68
v = Sqrt (11319.68) = 106.39 ft/s ≈ 32.43 m/s moving downwards.
5 0
3 years ago
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