When 597 J of heat are added to a gas, it expands. Its internal energy increases by 318 J. How much work does the gas do? (Unit=
1 answer:
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Answer:
The distance traveled is 109.58 m
Explanation:
The Speed of sound in air = 344 m/s
Let the time in which the dime dropped by the man reach an impact in the well = t₁
Let the time in which the sound travel from the well to the man = t₂
Then
1/2× 10 × t₁² = 344 × t₂ which gives;
5 × t₁² = 344 × t₂.........................(1)
Also the total time before the man heard the dime = t₁ + t₂ = 5
Therefore;
t₂ = 5 - t₁
Substituting the value of t₂ in equation (1), we have;
5 × t₁² = 344 × (5 - t₁)
5·t₁² + 344·t₁ - 1720 = 0
Using the quadratic formula, we have;
Which gives;
t₁ = 4.68 s or -73.48 s
Therefore, with the positive value for t₁ = 4.68 s, we have
The distance = 1/2× 10 × 4.68² = 109.58 m
The distance traveled = 109.58 m.
Answers:
a)
b)
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:
<h3><u>Snowball 1:</u></h3><u>x-component: </u>
(1)
Where:
is the initial speed of snowball 1 (and snowball 2, as well)
is the angle for snowball 1
is the time since the snowball 1 is thrown until it hits the opponent
<u>y-component: </u>
(2)
Where:
is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)
is the final height of the snowball 1
is the acceleration due gravity (always directed downwards)
<u>x-component: </u>
(3)
Where:
is the angle for snowball 2
is the time since the snowball 2 is thrown until it hits the opponent
<u>y-component: </u>
(4)
Having this clear, let's begin with the answers:
<h2>a) Angle for snowball 2</h2>Firstly, we have to isolate from (2):
(5)
(6)
Substituting (6) in (1):
(7)
Rewritting (7) and knowing :
(8)
(9)
(10) This is the point at which snowball 1 hits and snowball 2 should hit, too.
With this in mind, we have to isolate from (4) and substitute it on (3):
(11)
(12)
Rewritting (12):
(13)
Finding :
(14)
(15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle
.
<h2>b) Time difference between both snowballs</h2>
Now we will find the value of and
from (6) and (11), respectively:
(16)
(17)
(18)
(19)
Since snowball 1 was thrown before snowball 2, we have:
(20)
Finding the time difference between both:
(21)
Finally: