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AnnyKZ [126]
3 years ago
7

Red Riding Hood sat down to rest and placed her 1.20-kg basket of goodies beside her. A wolf came along, spotted the basket, and

began to pull on the handle with a force of 6.40 N at an angle of 25° with respect to vertical. Red was not going to let go easily, so she pulled on the handle with a force of 12.5 N. If the net force on the basket is straight up, at what angle was Red Riding Hood pulling from the vertical? (show work)
Physics
1 answer:
anygoal [31]3 years ago
3 0

Answer:

  117.6°

Explanation:

The vertical component of a force directed at some angle α from the vertical is ...

  F·cos(α)

We want the vertical components of the wolf's force (Fw) and Red's force (Fr) to total zero. So for some angle from vertical α, Red's force will satisfy ...

  Fw·cos(25°) + Fr·cos(α) = 0

  cos(α) = -Fw/Fr·cos(25°) ≈ -(6.4 N)/(12.5 N)·0.906308 ≈ -0.464030

  α ≈ arccos(-0.464030) ≈ 117.6°

Red was pulling at an angle of about 117.6° from the vertical.

_____

<em>Additional comment</em>

That's about 27.6° below the horizontal.

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A 72-kg skydiver is falling from 10000 feet. At an instant during the fall, the skydiver
MissTica

Answer:

Approximately 2.31\; \rm m \cdot s^{-2} (assuming that the acceleration due to gravity isg = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Assuming that g = 9.81\; \rm m \cdot s^{-2} the weight on this 72-kg skydiver would be W = m \cdot g = 72 \; \rm kg \times 9.81\; \rm m \cdot s^{-2} = 706.32\; \rm N (points downwards.)

Air resistance is supposed to act in the opposite direction of the motion. Since this skydiver is moving downwards, the air resistance on the skydiver would point upwards.

Therefore, the net force on this skydiver should be the difference between the weight and the air resistance on the skydiver:

\begin{aligned}F(\text{net force}) &= W - F(\text{air resistance})\\ &= 706.32\; \rm N - 540\; \rm N =166.32\; \rm N \end{aligned}.

Apply Newton's Second Law of motion to find the acceleration of this skydiver:

\begin{aligned}a &= \frac{F(\text{net force})}{m} \\ &= \frac{166.32\; \rm N}{72\; \rm kg} = 2.31\; \rm m \cdot s^{-2} \end{aligned}.

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3 years ago
Help. I need to find the degree using the sine law.
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4 0
3 years ago
If you saw a waxing gibbous moon what phase would you expect one week later?
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3 years ago
Explain why you can only change one variable in an experiment
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5 0
3 years ago
For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo
Snezhnost [94]

Answer:

114.86%

Explanation:

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Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

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Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

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Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

5 0
3 years ago
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