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Mila [183]
3 years ago
13

If it takes 42 minutes to load 3 1/2 trucks, how many minutes will it take to load 6 1/2 trucks?

Physics
1 answer:
Kruka [31]3 years ago
6 0
If it takes 42 minutes to load 3 1/2 trucks, then it takes 42 / 3.5 = 12 minutes to load a single truck. Multiplying the number of minutes per truck by 6.5 yields the time it will take to load. In this case 12 x 6.5 = 78 minutes would be required to load 6 1/2 trucks. The formula for the time taken is t = 12n, where t is the time in minutes and n is the number of trucks. 
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Find the height of the coconut from the ground at the moment the arrow hit it.
Wewaii [24]

Answer:

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Suppose a soccer ball is kicked from the ground at an angle 20.0º above the horizontal at 8.00 m/s. The y-velocity is determined
julia-pushkina [17]

initial speed of the ball by which it is kicked is 8 m/s

now the angle at which it is kicked is 20 degree

here we will have

v_y = v sin20

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now we will have

\Delta y = v_y t + \frac{1}{2}at^2

so if the ball again land on the ground at same level then we have

\Delta y = 0

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8 0
4 years ago
What would be an example of a chemical change that took a long time to occur
weqwewe [10]
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3 0
4 years ago
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9. How much work is done when a 15kg box is lifted to a height of 2 meters?
torisob [31]

Answer: W = 294 J

Explanation: Solution:

Work is expressed as the product of force and the distance of the object.

W = Fd        where F = mg

W= Fd

   = mg d

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   = 294 J

5 0
3 years ago
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.
Mice21 [21]

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.

4 0
3 years ago
Read 2 more answers
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