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coldgirl [10]
3 years ago
9

When cleaning up a local beach, students found many different particles that were in the water that affected the shoreline, like

litter and gasoline. Which of the following describe how these items would need to separated from the water? Why?
A. Both gasoline and litter would need to be chemically separated from the water, because both form new bonds with the water.


B. Litter would need to be physically separated with water and gasoline would need to be chemically separated from water, because both form a mixture with the water.


C. Both gasoline and litter would need to be chemically separated from the water, because neither bonds with the water.


D. Both gasoline and litter would need to be physically separated from the water, because neither bonds with the water.


(p.s. Do teachers know about this? like fr)
Physics
1 answer:
Wewaii [24]3 years ago
4 0

The correct answer is D,

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Answer:

A) ΔU = 3.9 × 10^(10) J

B) v = 8420.75 m/s

Explanation:

We are given;

Potential Difference; V = 1.3 × 10^(9) V

Charge; Q = 30 C

A) Formula for change in energy of transferred charge is given as;

ΔU = QV

Plugging in the relevant values gives;

ΔU = 30 × 1.3 × 10^(9)

ΔU = 3.9 × 10^(10) J

B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.

This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.

Thus;

P.E = K.E

ΔU = ½mv²

Where v is final velocity.

Plugging in the relevant values;

3.9 × 10^(10) = ½ × 1100 × v²

v² = [7.8 × 10^(8)]/11

v² = 70909090.9090909

v = √70909090.9090909

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4 0
3 years ago
A pilot heads her jet due east. The jet has a speed of 425 mi/h relative to the air (in other words, if the air were still, the
Elza [17]

Answer:

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Sketching out the problem given, by using straight lines to represent each of the vectors, we will have a right angled triangle as shown below.

The solution can be obtained by applying Pythagoras theorem to

resolve the vectors.

Velocity of jet plane = 425 mi/hr

velocity of air = 40 mi/hr

Resultant of the vectors =\sqrt[]{425^{2}+40^{2}}=426.87 mi/hr

Vector direction =tan^{-1}(\frac{40}{425})= 5.36 degrees

hence the velocity is 426.87 mi/hr in a direction 5.36 degrees inclined Northward

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77julia77 [94]

Answer:

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\lambda=\dfrac{\lambda_0}{\sqrt{\varepsilon_r}}\\\Rightarrow \lambda=\dfrac{0.61}{\sqrt{1.44}}\\\Rightarrow \lambda=0.5083\ \mu m

The wavelength lies in the range of green light.

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3 years ago
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Answer:

d. none of these

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From the given information:

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FOR Sir Rodney;

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

S = (3\times t ) + \dfrac{1}{2} \times 0 \times t^2  \\ \\  S =3t---  (2)

Equating both equations together; we have:

0.3t² = 3t

0.3t² - 3t = 0

0.3t(t - 10) = 0

If Percival's position at rest = 0

Then; t = 10 s.

8 0
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