A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the
1 answer:
Answer:
+16 J
Explanation:
We can solve the problem by using the 1st law of thermodynamics:
where
is the change of the internal energy of the system
Q is the heat (positive if supplied to the system, negative if dissipated by the system)
W is the work done (positive if done by the system, negative if done by the surroundings on the system)
In this case we have:
Q = -12 J is the heat dissipated by the system
W = -28 J is the work done ON the system
Substituting into the equation, we find the change in internal energy of the system:
You might be interested in
Answer:
The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s
Explanation:
The given parameters of the motion of the branch are;
The height from which the branch is thrown = 3.00 m
The horizontal distance the branch lands from where it was thrown, x = 8.00 m
The direction in which the branch is thrown = Horizontally
Therefore, the initial vertical velocity of the branch, = 0 m/s
The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;
s = ·t + 1/2·g·t²
Where;
= 0 m/s
s = The initial height of the object = 3.00 m
g = The acceleration due to gravity = 9.8 m/s²
∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²
t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246
The horizontal distance covered before the branch touches the ground, x = 8.00 m
Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;
uₓ = x/t = 8.00 m/((√30)/7 s)
Using a graphing calculator, we get;
uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s
The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.
(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².
(b) The time when the velocity of the turtle is zero is 3.2 s.
(c) The time taken for the turtle to return to its starting point is 6.4 s.
(d) The time taken for the turtle to travel 30 cm is 0.08 s.
<h3>Initial velocity of the turtle</h3>The initial velocity of the turtle is calculated as follows;
The initial acceleration of the turtle is calculated as follows;
x(t) = 5 + 4t - 0.625t²
x(0) = 5 cm
<h3>Time when the velocity becomes zero</h3>v(t) = 4 - 1.25t
0 = 4 - 1.25t
1.25t = 4
t = 4/1.25
t = 3.2 s
<h3>Time taken to return to starting point</h3>The total distance traveled is calculated as follows
v² = u² + 2ad
0 = (4)² + 2(-1.25)d
0 = 16 - 2.5d
2.5d = 16
d = 16/2.5
d = 6.4 m
Time to travel the given distance;
d = ut + ¹/₂at²
6.4 = (4)t + ¹/₂(-1.25)t²
6.4 = 4t - 0.625t²
0.625t² - 4t + 6.4 = 0
solve the quadratic equation using formula method;
t = 3.2 s
The time travel the distance two times, = 2 x 3.2 s = 6.4 s
<h3>Time taken for the turtle to travel 30 cm</h3>d = ut + ¹/₂at²
0.3 = (4)t + ¹/₂(-1.25)t²
0.3 = 4t - 0.625t²
0.625t² - 4t + 0.3 = 0
solve the quadratic equation using formula method;
t = 0.08 s
Learn more about velocity here: brainly.com/question/6504879