Why is styrofoam not a mineral?
2 answers:
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Answer:
t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s
Explanation:
Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso
y = y₀ + t – ½ g t²
en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)
0 = y₀ – ½ g t²
t= √ (2 y₀/g)
calculemos
t= √ ( 2 12 / 9,8)
t= 1,56 s
El alcance del proyectil es la distancia horizontal recorrida
x = v₀ₓ t
x = 80 1,56
x= 124,8 m
La velocidad de impacto cuando toca el suelo
vx = v₀ₓ = 80 ms
= v_{oy} – gt
v_{y} = - 9,8 1,56
v_{y} = - 15,288 m/s
la velocidad es
v = (80 i^ - 15,288 j ) m/s
Traducttion
This is a projectile launching exercise, let's start by finding the time it takes to reach the ground
y = y₀ + v_{oy} t - ½ g t²
in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( = 0)
0 =y₀I - ½ g t²
t = √ (2y₀ / g)
let's calculate
t = √ (2 12 / 9.8)
t = 1.56 s
Projectile range is the horizontal distance traveled
x = v₀ₓ t
x = 80 1.56
x = 124.8 m
Impact speed when it hits the ground
vₓ = v₀ₓ = 80 ms
v_{y} = v_{oy} - gt
v_{y} = - 9.8 1.56
v_{y} = - 15,288 m / s
the speed is
v = (80 i ^ - 15,288 j) m / s
a) The impulse is 76.5 Ns
b) The average force is 546.4 N
c) The final speed is 31.5 m/s
Explanation:
a)
The impulse exerted on an object is defined as
where
F is the magnitude of the force exerted on the object
is the time interval during which the force is applied
If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.
Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are
and whose height is
Therefore, the area (and the impulse) is
b)
In this problem, the force applied is not constant. However, we can rewrite the impulse also as
where
is the average force exerted during the whole time
In this problem we have
J = 76.5 Ns is the impulse (calculated in part a)
is the time interval
Solving for the average force, we find
c)
According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:
where
m is the mass of the object
v is the final velocity
u is the initial velocity
In this problem, we have
J = 76.5 Ns
m = 3.0 kg is the mass
u = 6.0 m/s is the initial velocity
Solving for v, we find the final velocity (and speed):
Learn more about impulse and momentum:
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Answer:
I)
II)The difference from the value obtained on part I is: 2000-1385.67 =614.33
The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.
Explanation:
From the problem we have the molar mass given of water vapor and at constant volume condition. It's important to say that the vapour molecules have 3 transitionsl and 3 rotational degrees of freedom and the rotational motion no contribution.
Part I
Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational and three rotational degrees of freedom and that vibrational motion does not contribute. The molar mass of water is 18.0 g/mol=0.018kg/mol.
Let the molar heat capacity at constant volume and this amount represent the quantity of heat absorbed by mole.
Let the specific heat capcity this value represent the heat capacity aboserbed by mass.
For the problem we have a total of 6 degrees of freedom and from the thoery we know that for each degree of freedom the molar heat capacity at constant volume is given by so the total for the 6 degrees of freedom would be:
And by definition we know that the specific heat capacity is defined:
If we replace all the values we have:
So on this case the specific heat capacity with constant volume and with three translational and three rotational degrees of freedom is
Part II
The actual specific heat of water vapor at low pressures is about 2000 J/(kg * K). Compare this with your calculation.
The difference from the value obtained on part I is: 2000-1385.67 =614.33
The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.