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lorasvet [3.4K]
3 years ago
12

Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is r

eleased
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

<em>171.5m</em>

Explanation:

The velocity of sound in water = 343m/s

Time taken = 1.00secs

using the formula to calculate the distance

2x = vt

x is the distance

v is the speed of sound

t is the time

x = vt/2

x = 343(1)/2

x = 171.5m

<em>hence their separation 1.00 s after the second object is released is 171.5m</em>

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By using the formula:

• K.E = ½mv²

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A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how bi
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Answer:

a) F₁₂₀ = 1.34 pa A  , b)  F₂₀ = 0.746 pa A

Explanation:

Part. A .    The definition of pressure is

         P = F / A

As the air can approach an ideal gas we can use the ideal gas equation

        P V = n R T

Let's write this equation for two temperatures

       P₁ V = n R T₁

       P₂2 V = n R T₂

       P₁ / P₂ = T₁ / T₂

point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂

       P₂ = P₁ T₂ / T₁

       P₂ = pa 393/293

       P₂ = 1.34 pa

We calculate the strength

       P₂ = F₁₂₀ / A

       F₁₂₀ = 1.34 pa A

Part B

In this case the data is

Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure

        P₁ / P₂ = T₁ / T₂

       P₂ = P₁ T₂ / T₁

       P₂ = pa 293/393

       P₂ = 0.746 pa

Let's calculate the force (F20), from this point

      F₂₀ / A = 0.746 pa

     F₂₀ = 0.746 pa A

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