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musickatia [10]

4 years ago

7

A 3-liter container has a pressure of 4 atmospheres. The container is sent underground, with resulting compression into 2 L. App

lying Boyle's Law, what will the new pressure be? choices: 2.3 atm 8 atm 6 atm 1.5 atm

1 answer:

monitta4 years ago

4 0

Answer:

6 atm

Explanation:

PV = PV

(4 atm) (3 L) = P (2 L)

P = 6 atm

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The kinetic energies of particles in a sample of matter are increasing. This sample is most likely

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The sample is most likely gaining thermal energy

Explanation:

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3 years ago

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Myopia is a condition of the eye where the crystalline lens focuses the light rays to a position between the lens and the retina

aliya0001 [1]

Answer: concave lens

Explanation:

Myopia is a condition of the eye where someone can only see near distant object clearly but not far distant object.

Myopia is corrected using concave lens (diverging) in order to diverge the rays entering the eye thereby allowing the rays to be focused properly on the retina.

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3 years ago

Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Poi

djyliett [7]

Answer:

A) The particle will accelerate in the direction of point C.

Explanation:

As we know that

potential at points A, B,C and D as V_A, V_B, V_C, V_D and it is clear from the question that

V_A>V_B>V_C

And we know that flow is always from higher to lower potential (for positive charge due to positive potential energy).

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Hence, the correct option is A.

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3 years ago

Which statement about moons is true?

kotegsom [21]

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4 years ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago