Answer:
a) The speed is 61.42 m/s
b) The drag force is 10.32 N
Explanation:
a) The Reynold´s number for the model and prototype is:
Equaling both Reynold's number:
Clearing Vm:
b) The drag force is:
When a 1.0-kilogram cart moving with a speed of 0.50 meter per second on a horizontal surface collides with a second 1.0-kilogra
1 answer:
The speed of the combined carts after the collision is 0.25 m/s
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system must be conserved before and after the collision, so we can write:
where:
is the mass of the first cart
is the initial velocity of the first cart
is the mass of the second cart
is the initial velocity of the second cart
is the final combined velocity of the two carts
Re-arranging the equation and substituting the values, we find: the final velocity:
Learn more about momentum:
#LearnwithBrainly
You might be interested in
Answer:
a) F = (m / t₀) 95.33
b) θ = 70.5º
Explanation:
This is a projectile launch, as they indicate the horizontal distance this is the range of the body, let's use the expression for the range of the projectiles
R = v₀² sin 2θ / g
v₀² sin 2θ = R g
Where the range is 550.46 m
They also indicate the time that the air must remain, so this time is twice the time until reaching the maximum height.
v_{y} = - g t
At the maximum height v_{y} = 0 and the initial speed on the axis and we can find it with trigonometry
sin θ = v_{oy} / v_{o}
v_{oy} = v_{o} sin θ
v_{o} sin θ = g t
Let's write the two equations
v_{oy}² sin 2θ = g R
v_{o} sin θ = g t
We solve our accusation system
(G t / sin θ) 2 sin 2θ = g R
g t² sin 2θ = R sin θ
Let's use the trigonometric relationship
sin 2θ = 2 sin θ cos θ
We substitute
g t² (2 sin θ cos θ) = R sin θ
Cos tea = R / 2 g t²
θ = cos⁻¹ (R / 2g t²)
Let's calculate
θ = cos⁻¹ (550.46 / (2 9.8 9.17² ))
θ = 70.5º
a) Force can be Newton's second law
On the x axis the speed is constant so the force on the axis is zero
In the y axis the acceleration we have is the acceleration of gravity, so the force that acts throughout the journey is the weight of the body.
To place the body in the air from the rest we can use the equation of the impulse
F t = Δp = m v - m v₀
As kick from rest v₀ = 0
Let's find the speed of the body
= g t
v_{o} = g t / sin 70.5
v_{o} = 9.8 9.17 / sin 70.5
v_{o} = 95.33 m / s
To encocorate the force we must suppose a firing time, which in general is very short, suppose that this time is to
F = m v_{o} / t₀
F = (m / t₀) 95.33
This is the outside that should be applied, as an example suppose a body of mass 1 kg⁵ ( m = 1 kg) and a trip time to = 0.1 s
F = (1 / 0.1) 95.33
F = 953.3 N