Can a runner on a track be accelerating even if she is running at a constant speed why or why not
1 answer:
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To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.
In turn, we will resort to the application of Newton's second law.
PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,
Where,
X = Desplazamiento
V = Velocity
t = Time
In this case there is no initial displacement or initial velocity, therefore
Clearing for time,
PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:
F = ma
Where,
m=mass
a = acceleration
Acceleration can also be written as,
Then
Negative symbol is because the force is opposite of the direction of moton.
PART C) Acceleration through kinematics equation is defined as
The gravity is equal to 0.8, then the acceleration is
Answer:
The temperature of the steam during the heat rejection process is 42.5°C
Explanation:
Given the data in the question;
the maximum temperature T in the cycle is twice the minimum absolute temperature T
in the cycle
T = 0.5T
now, we find the efficiency of the Carnot cycle engine
η = 1 - T
/T
η = 1 - T
/0.5T
η = 0.5
the efficiency of the Carnot heat engine can be expressed as;
η = 1 - W
/Q
where W is net work done, Q
is is the heat supplied
we substitute
0.5 = 60 / Q
Q = 60 / 0.5
Q = 120 kJ
Now, we apply the first law of thermodynamics to the system
W = Q
- Q
60 = 120 - Q
Q = 60 kJ
now, the amount of heat rejection per kg of steam is;
q = Q
/m
we substitute
q = 60/0.025
q = 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q = h
= 2400 kJ/kg
now, at h = 2400 kJ/kg from saturated water tables;
T = 40 + ( 45 - 40 ) (
)
T = 40 + (5) × (0.5)
T = 40 + 2.5
T = 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C
Answer:
d= 64.1 km θ = 36.4º
Explanation:
a) In order to find the plane's straight-line distance from the starting point, we need to know the coordinates of the final and initial position of the plane, so we can find the total displacement, as the difference between the final and initial position.
If we choose to put our origin at the initial point of trajectory, we have that (x₀, y₀) = (0, 0)
In order to find the position of the plane after finishing the flight, we need to find its final coordinates (x₁, y₁).
In order to get x₁, we need to add the x-coordinate after flying 45º north of east, and the Δx after completing the flight in a direction 25º of east, that we can find applying trigonometry, as follows:
x₁ = 37.0 km * cos 45º + 28.0 km* cos 25º = 51.6 Km
Appying the same considerations for the y-coordinate, we have:
y₁ = 37.0 km * sin 45º + 28.0 km* sin 25º = 38.0 km
Now, as the initial position coincides with the origin, the distance in a straight line from this point to the origin, is just the hypotenuse of the triangle determined by the coordinates (x₁, y₁) and (0,0), as follows:
The geographic direction of the displacement vector (which coincides in magnitude with the distance we have just found), is just the angle that this distance forms with the east axis, that we can find getting the tangent of this angle as follows:
tg θ =
⇒ θ = 36.4º North of East (counterclockwise from the east axis).