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3 years ago

Liquid droplets that fall from the sky are? A) rain B)snow C) sleet D) hail

Physics

2 answers:

Gre4nikov [31]3 years ago

8 0

The correct answer is A) Rain.

Answer A: Rain is correct because rain is liquid that falls from the sky.

Answer B: Snow is incorrect because it's frozen rain that falls from the sky.

Answer C: Sleet is incorrect because it's partially frozen, not entirely liquid.

Answer D: Hail is incorrect because hail is ice that falls from the sky.

alukav5142 [94]3 years ago

6 0

The answer is a)rain

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m_a_m_a [10]

The answer is b. The crest.

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3 years ago

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Gasoline burns inside a car’s engine. how does this fuel enable a car to move?

Svetllana [295]

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8 0

3 years ago

The temperature of a balloon is lowered with liquid nitrogen. The balloon appears to deflate because the A. nitrogen has burst t

Elina [12.6K]

The temperature of a balloon is lowered with liquid nitrogen. The balloon appears to deflate because the gas in the balloon lost energy and condensed. The answer is letter D.

3 0

3 years ago

How to solve it? Three capacitors with capacities of 600 pF, 300 pF, 200 pF are connected in series. The 60 V voltage is applied

adell [148]

Answer:

1. Voltage across 600 pF is 10 V.

2. Voltage across 300 pF is 20 V.

3. Voltage across 200 pF is 30 V.

Explanation:

We'll begin by calculating the total capacitance of capacitor. This can be obtained as follow:

Capicitance 1 (C₁) = 600 pF

Capicitance 2 (C₂) = 300 pF

Capicitance 3 (C₃) = 200 pF

Total capacitance (Cₜ) =?

1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

1/Cₜ = 1/600 + 1/300 + 1/200

1/Cₜ = 1 + 2 + 3 / 600

1/Cₜ = 6/600

1/Cₜ = 1/100

Cₜ = 100 pF

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Thus, 100 pF is equivalent to 1×10¯¹⁰ F.

Next, we shall determine the charge. This can be obtained as follow:

Voltage (V) = 60 V

Capicitance (C) = 1×10¯¹⁰ F

Charge (Q) =?

Q = CV

Q = 60 × 1×10¯¹⁰ F

Q = 6×10¯⁹ C

1. Determination of the voltage across 600 pF.

Capicitance 1 (C₁) = 600 pF = 6×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 1 (V₁) =?

Q = C₁V₁

6×10¯⁹ = 6×10¯¹⁰ × V₁

Divide both side by 6×10¯¹⁰

V₁ = 6×10¯⁹ / 6×10¯¹⁰

V₁ = 10 V

2. Determination of the voltage across 300 pF.

Capicitance 2 (C₂) = 300 pF = 3×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 2 (V₂) =?

Q = C₂V₂

6×10¯⁹ = 3×10¯¹⁰ × V₂

Divide both side by 3×10¯¹⁰

V₂ = 6×10¯⁹ / 3×10¯¹⁰

V₂ = 20 V

3. Determination of the voltage across 200 pF.

Capicitance 3 (C₃) = 200 pF = 2×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 3 (V₃) =?

Q = C₃V₃

6×10¯⁹ = 2×10¯¹⁰ × V₃

Divide both side by 2×10¯¹⁰

V₃ = 6×10¯⁹ / 2×10¯¹⁰

V₃ = 30 V

7 0

3 years ago

Which statement best define percent error? check all that apply.

Art [367]

Percent error is the difference between the experimental value and theoretical value and measures the accuracy of the result found. The larger the error, lesser is the accuracy and vice versa.

Solution:

It is a mathematical way of showing accuracy

The higher the percent error, the less accurate the data set,

6 0

1 year ago