If a stone is thrown vertically upward from the surface of the moon with a velocity of 13 m/s, its height (in meters) after t seconds is h = 13t - 0.83t2. (Round the answers to two decimal places.)
If a stone is thrown vertically upward from the surface of the moon with a velocity of 13 m/s, its height (in meters) after t seconds is h = 13t - 0.83t2. (Round the answers to two decimal places.)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem Statement:**
If a stone is thrown vertically upward from the surface of the moon with a velocity of 13 m/s, its height (in meters) after \( t \) seconds is given by the formula \( h = 13t - 0.83t^2 \). (Round the answers to two decimal places.)
**Questions:**
(a) What is the velocity of the stone after 4 seconds?
- Answer Box: \( v = \, \) m/s (Incorrect submission indicated by an "X")
(b) What is the velocity of the stone after it has risen 24 meters?
- Answer Box: \( v = 11.22546294 \) m/s (Incorrect submission indicated by an "X")
**Explanation:**
The problem involves finding the velocity of the stone at specific instances by using the given height formula. The velocity can be found by differentiating the height function \( h(t) = 13t - 0.83t^2 \) with respect to time \( t \).
The derivative will give the velocity function:
\[
v(t) = \frac{dh}{dt} = 13 - 2 \times 0.83 \times t = 13 - 1.66t
\]
For part (a), substitute \( t = 4 \) seconds into the velocity function to find the velocity at that time.
For part (b), solve the equation \( h(t) = 24 \) to find the time \( t \) when the stone has risen 24 meters. Then, substitute this time back into the velocity function to find the velocity at that height.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78d57d79-5bdd-470b-8043-f6ada2e59fe2%2F2aa77fe1-e477-4520-8a7a-a9368a85e0f0%2Fiq2ifsp_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
If a stone is thrown vertically upward from the surface of the moon with a velocity of 13 m/s, its height (in meters) after \( t \) seconds is given by the formula \( h = 13t - 0.83t^2 \). (Round the answers to two decimal places.)
**Questions:**
(a) What is the velocity of the stone after 4 seconds?
- Answer Box: \( v = \, \) m/s (Incorrect submission indicated by an "X")
(b) What is the velocity of the stone after it has risen 24 meters?
- Answer Box: \( v = 11.22546294 \) m/s (Incorrect submission indicated by an "X")
**Explanation:**
The problem involves finding the velocity of the stone at specific instances by using the given height formula. The velocity can be found by differentiating the height function \( h(t) = 13t - 0.83t^2 \) with respect to time \( t \).
The derivative will give the velocity function:
\[
v(t) = \frac{dh}{dt} = 13 - 2 \times 0.83 \times t = 13 - 1.66t
\]
For part (a), substitute \( t = 4 \) seconds into the velocity function to find the velocity at that time.
For part (b), solve the equation \( h(t) = 24 \) to find the time \( t \) when the stone has risen 24 meters. Then, substitute this time back into the velocity function to find the velocity at that height.
![The gas law for an ideal gas at absolute temperature \( T \) (in kelvins), pressure \( P \) (in atmospheres), and volume \( V \) (in liters) is given by the equation \( PV = nRT \), where \( n \) is the number of moles of the gas and \( R = 0.0821 \) is the gas constant.
Suppose that, at a certain instant, \( P = 8.0 \) atm and is increasing at a rate of \( 0.13 \) atm/min, while \( V = 12 \) L and is decreasing at a rate of \( 0.16 \) L/min. Find the rate of change of \( T \) with respect to time at that instant if \( n = 10 \) mol.
Round your answer to four decimal places.
\[
\frac{dT}{dt} = \, \text{K/min}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78d57d79-5bdd-470b-8043-f6ada2e59fe2%2F2aa77fe1-e477-4520-8a7a-a9368a85e0f0%2F7xu9v2_processed.png&w=3840&q=75)
Transcribed Image Text:The gas law for an ideal gas at absolute temperature \( T \) (in kelvins), pressure \( P \) (in atmospheres), and volume \( V \) (in liters) is given by the equation \( PV = nRT \), where \( n \) is the number of moles of the gas and \( R = 0.0821 \) is the gas constant.
Suppose that, at a certain instant, \( P = 8.0 \) atm and is increasing at a rate of \( 0.13 \) atm/min, while \( V = 12 \) L and is decreasing at a rate of \( 0.16 \) L/min. Find the rate of change of \( T \) with respect to time at that instant if \( n = 10 \) mol.
Round your answer to four decimal places.
\[
\frac{dT}{dt} = \, \text{K/min}
\]
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