Answer:
If I'm not working I think the answer is C.
La respuesta es "Un avion que vuela al norte con rapidez constante y altitud constante".
Para que la fuerza neta sea 0, la aceleracion debe ser 0, para esto la velocidad debe ser constante.
Para que la velocidad sea constante el objecto debe estar moviendo con rapidez (magnitud de la velocidad) constante y sin cambiar direccion; ya que la velocidad es un vector asi es que depende en magnitud y direccion.
En las demas opciones la magnitud de la velocidad (rapidez) cambia y/o la direccion.
Explanation:
It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :


p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.
Answer:
A) θ = 13.1º , B) E
Explanation:
A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.
In this reference system the only force that we must decompose is the Normal one, let's use trigonometry
sin θ = Nₓ / N
cos θ =
/ N
Nₓ = N sin θ
Ny = N cos θ
x-axis (radial)
Nₓ = m a
where the acceleration is centripetal
a = v² / R
we substitute
-N sin θ = -m v² / R (1)
the negative sign indicates that the force and acceleration towards the center of the circle
y-axis (Vertical)
Ny - W = 0
N cos θ = mg
N = mg / cos θ
we substitute in 1
mg / cos θ sin θ = m v² / R
g tan θ = v² / R
θ = tan⁻¹ (v² / gR)
we calculate
θ = tan⁻¹ (25² / 9.8 274)
θ = 13.1º
B) when comparing the equations the correct one is E
<span>Answer: 1600 J
Explanation:
1) Data:
a) ideal gas: ⇒ pV = nRT and work = ∫ pdV
b) slowly compressed ⇒ constant temperature and not heat exchange
c) pressure: p = 2 atm
d) intitial volume: Vi = 10 liters
e) final volumen: Vf = 2 liters.
f) then the
volume of the gas is held constant ⇒ not work in this stage.
g) calculate the work done on the gas: W = ?
2) Equation
W = ∫pdV
3) Solution:
Since p = constant, W = p ∫dV = p ΔV = p (Vf - Vi)
p = 2 atm × 1.0 ×10⁵ Pa / atm = 200.000 Pa
Vi = 10 liter × 0.001 m³ ./ liter = 0.01 m³
Vf = 2 liter × 0.001 m³ / liter = 0.002 m³
W = 200.000 Pa × (0.002 m³ - 0.01m³) = - 1.600 J.
The negative sign means the work is done over the system.
That is all the work in the system because at the second stage the volume is held constant.
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