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9966 [12]
3 years ago
14

Complete the statements using data from Table A of your Student Guide. The speed of the cart after 8 seconds of Low fan speed is

cm/s. The speed of the cart after 3 seconds of Medium fan speed is cm/s. The speed of the cart after 6 seconds of High fan speed is cm/s.
Physics
2 answers:
Colt1911 [192]3 years ago
8 2

Answer:

The speed of the cart after 3 seconds of Low fan speed is 54.0 cm/s

The speed of the cart after 5 seconds of Medium fan speed is 120.0 cm/s

The speed of the cart after 2 seconds of High fan speed is 64.0 cm/s

Explanation:

just did it on edge

finlep [7]3 years ago
4 0

Answer:

The speed of the cart after 8 seconds of Low fan speed is  72.0 cm/s

The speed of the cart after 3 seconds of Medium fan speed is   36.0 cm/s

The speed of the cart after 6 seconds of High fan speed is  96.0 cm/s

Explanation:

took the test on edgenuity

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4 years ago
Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,
EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

          F₁ = 2.50 kN

Tide

         cos 30 = F₂ₓ / F₂

         sin 30 = F_{2y} / F₂

          F₂ₓ = F₂ cos 30

         F_{2y} = F₂ sin 30

         F₂ₓ = 1.20cos 30 = 1.039 kN

         F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

        Fₓ = F₁ₓ + F₂ₓ

        Fₓ = 2.50 +1.039

        Fₓ = 3,539 kN

        F_y = F_{2y}

        F_y = 0.600

to find the vector we use the Pythagorean theorem

         F = \sqrt{F_x^2 +F_y^2}

         F = \sqrt{ 3.539^2 + 0.600^2 }

         F = 3,589 kN

the address is

         tan θ = F_y / Fₓ

         θ = tan⁻¹ \frac{F_y}{F_x}

         θ = tan⁻¹  \frac{0.6}{3.539}0.6 / 3.539

         θ = 9.6º

the resultant force to two significant figures is

         F = 3.6 kN

the direction is 9.6º to the North - East

7 0
3 years ago
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