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Amanda [17]
2 years ago
13

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

uation PV = C, where C is a constant. Suppose that at a certain instant the volume is 1000 cm3, the pressure is 80 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?
Mathematics
1 answer:
krek1111 [17]2 years ago
5 0

Answer:

The volume is decreasing at a rate of 250 cm^3/min

Step-by-step explanation:

From Boyle's law, the relationship between pressure (P) and volume (V) is inverse in which increase in one quantity (P) leads to a corresponding decrease in the other quantity (V).

Pressure at a certain instant (time) = 80 kPa

Rate at which pressure is increasing = 20 kPa/min

Time required for this increase = pressure ÷ rate = 80 ÷ 20 = 4 min

Volume = 1000 cm^3

Rate at which volume is decreasing at the same time pressure is increasing = 1000 ÷ 4 = 250 cm^3/min

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An orange is shot up into the air with a catapult. The function h given by h(t)=15+60t-16t^2 models the orange’s height, in feet
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Complete question is:An orange is shot up into the air with a catapult. The function h given by h(t) = 15 + 60t - 16t² models the orange’s height, in feet, t seconds after it was launched.

Select all the true statements about the situation.

Options:

1. The domain of function h only contains values greater than or equal to 0.

2. The orange is at the same height 1 second after launch and 2 seconds after launch.

3. After 3 seconds, the orange has hit the ground.

4. The orange is 15 feet above the ground when it is launched.

5. The value t = 10 does not belong to the domain of h.

Answer:

Option 1 - True

Option 2 - False

Option 3 - False

Option 4 - True

Option 5 - True

Step-by-step explanation:

Looking at the options,

-The domain is the x or t value and it is time. Now, time can only be positive or greater than or equal to zero i.e. t ≥ 0. Thus, option 1 is true.

- At t = 1 second;

h = 15 + 60(1) - 16(1)²

h = 15 + 60 - 16 = 59 ft

Also, at t = 2 seconds;

h = 15 + 60(2) - 16(2)²

h = 15 + 120 - 64 = 71 ft

So,value of h is not the same at t = 1 and t = 2. Thus, option 2 is not true.

- The orange will hit the ground when h(t) = 0.

However, at t = 3;

h(t) = 15 + 60(3) - 16(3)²

h(t) = 51 ft

h(t) is not equal to zero at t = 3, so option 3 is false

- when the orange is launched it is at time t = 0.

Thus,

h(0) = 15 + 60(0) - 16(0)²

h(0) = 15 ft

So option 4 is true.

- At t=10, h(t) = 15 + 60(10) - 16(10)²

h(10) = -985

This means the orange would be below ground level and thus doesn't belong to the domain of h, so option 5 is true.

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