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ludmilkaskok [199]
3 years ago
9

A car's velocity as a function of time is given by vx(t)=α+βt2, where α=3.00m/s and β=0.100m/s3.

Physics
2 answers:
Jlenok [28]3 years ago
5 0

we are asked to draw a plot of the formula vx(t)=α+βt2 in which  α=3.00m/s and β=0.100m/s3. Using MS Excel, we can plot the points from t equal to zero to t equal to 5 seconds. The plot is attached in the file. 

Mice21 [21]3 years ago
4 0

1) Analyze the equation

Vₓ = α + β t²

Vₓ = 3.00 + 0.100 t²

That is a quadratic equation, so the graph is a parabola.

2) Therefore, take into account that the main points to draw a parabola are:

i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

3) Find the y-intercept (t = 0)

t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00

4) Find the x-intercepts (Vₓ = 0)

Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²

⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.

5) Concavity

Since, the coefficient of t² is positive, the parabola open upwards.

6) Vertex

It is the local minimum of the equation. You can find it by the first derivative

Vₓ' = 2×0.100 t = 0 ⇒ t = 0

Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s

⇒ vertex = (0,3.00)

7) The domain is given t ∈ [0,5.00]

8) You can also build a table with several points in the domain

t =0; Vₓ = 3.00 + 0.100 (0)² = 0

t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10

t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40

t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90

t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60

t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50

9) Range: Vₓ ∈ [ 3.00, 5.50]

10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.

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Answer:

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Explanation:

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To solve this problem, we use Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".

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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

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