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Artyom0805 [142]
4 years ago
9

When a space shuttle was launched, the astronauts aboard experienced and acceleration of 29.0 m/s. If one of the astronauts had

a mass of 69.0 kg, what net force in newtons did the astronaut experience?
Physics
1 answer:
ankoles [38]4 years ago
7 0

             Net force = (mass) · (acceleration)

                           = (69 kg) · (29 m/s²)

                           = (69 · 29) · (kg·m/s²)

                           =    2,001  Newtons upward
                         (about  450 pounds)
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Four forces act on bolt A as shown; F1 150N, F2 80N, F3 110N and F4 100N. Determine the magnitude and direction of the resultant
netineya [11]

Complete Question

The  complete question(reference (chegg)) is shown on the first uploaded image

Answer:

The magnitude of the resultant force is  F  =  199.64 \ N

The  direction of the resultant force is  \theta  =  4.1075^o from the horizontal plane

Explanation:

Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be  Fcos(\theta ) while if the force is  moving away from the angle  then the resolved force is  Fsin (\theta )

Now  from the diagram let resolve the forces to their horizontal component

    So

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Now  resolving these force into their vertical component can be mathematically evaluated as

         \sum  F_{y}  =  150 sin(30) - 100sin(15) -110 +80 cos(20)

         \sum  F_{y}  =  14.30

Now the resultant force is mathematically evaluated as

        F  =  \sqrt{F_x^2 + F_y^2}

substituting values

        F  =  \sqrt{199.128^2 + 14.3^2}

        F  =  199.64 \ N

The  direction of the resultant force is  evaluated as

       \theta  =  tan^{-1}[\frac{F_y}{F_x} ]

substituting values

       \theta  =  tan^{-1}[\frac{ 14.3}{199.128} ]

       \theta  =  4.1075^o from the horizontal plane

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Since point D is located at <u>negative -4</u> and point E is at <u>positive 5</u>, we have:

\overline{DE} = E - D = 5 - (-4) = 9

Hence, the segment of the line DE (\overline{DE}) is 9.

Knowing that point F partitions the line segment from D to E (\overline{DE}) into a <u>5:6 ratio</u>, its location would be:

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Learn more about segments here:

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<em />

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Answer:

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The frequency of the infrared light = v ≈ 3.156 × 10¹⁴ Hz.

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