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alexandr402 [8]
3 years ago
7

A pump contains 0.5 L of air at 203 kPa . You draw back on the piston of the pump, expanding the volume until the pressure reads

50.8 kPa . What is the new volume of air in the pump?
Physics
1 answer:
aksik [14]3 years ago
5 0

Answer:

V₂ = 2 L

Explanation:

given,

P₁ = 203 kPa

V₁ = 0.5 L

Now, volume is expanded

P₂ = 50.8 kPa

V₂ = ?

Using Boyle's law

 P₁V₁ = P₂ V₂

V_2 = \dfrac{P_1V_1}{P_2}

inserting all the values

V_2 = \dfrac{203\times 0.5}{50.8}

     V₂ = 2 L

Hence, new volume of air in the pump is equal to 2 L.

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Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

s_1 = v_1t = 1.6t

Similarly the distance (m) from your friend building to the restaurant:

s_2 = v_2t = 1.2t

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

s_1 + s_2 = 5b

1.6t + 1.2t = 5b

2.8t = 5b

t = 5b/2.8 = 25b/14

s_2 = 1.2t = 1.2(25b/14) = 2.143b

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

7 0
3 years ago
When the switch in the drawing is closed, the current in the coil increases to its equilibrium value. While the current is incre
Crazy boy [7]

Answer:

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Answer:

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Explanation:

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