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Phoenix [80]
2 years ago
13

A baby elephant is stuck in a mud hole. to help pull it out, game keepers use a rope to apply a force f with arrowa, as part a o

f the drawing shows. by itself, however, force f with arrowa is insufficient. therefore, two additional forces f with arrowb and f with arrowc are applied, as in part b of the drawing. each of these additional forces has the same magnitude f. the magnitude of the resultant force acting on the elephant in part b of the drawing is k times larger than that in part
a. find the ratio f fa when k = 2.10. (take θ = 18.0°.) f fa = 1.05 incorrect: your answer is incorrect.
Physics
1 answer:
Julli [10]2 years ago
3 0

The two forces should be equal therefore:

2.10 * Fa = Fa + 2 * F * cos 18

simplifying the right side:

2.10 * Fa = Fa + 1.902 * F

1.10 Fa = 1.902 F

<span>F / Fa = 0.578</span>

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A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>
  1. Determine the acceleration of the stuffed bear brainly.com/question/6268248
  2. Particle's speed and direction of motion brainly.com/question/2814900
  3. About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

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2 years ago
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A4 40 kg girl skates at 3.5 m/s one ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold
salantis [7]

Explanation:

Below is an attachment containing the solution.

8 0
2 years ago
A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?
yanalaym [24]

The average speed can be easily calculated by taking the ratio of distance and time. That is:

average speed = distance / time

 

so calculating:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

8 0
2 years ago
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For what value m of the clockwise couple will the horizontal component ax of the pin reaction at a be zero? if a couple of that
sergejj [24]
Thank you for posting your question here at brainly. Below is the answer:

sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M 
<span>- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) </span>

<span>sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M </span>
<span>- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) </span>

<span>Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) </span>
<span>= (2M + 424.91)/(5.745) </span>

<span>M = ((275*5.745) - 424.91)/2 </span>
<span>= 577.483 which rounds off to 577 </span>

<span>Is it maybe supposed to be Ay - Cy = 275</span>
8 0
2 years ago
You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car
gogolik [260]

B. The sound of the engine will get louder and the pitch higher.

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2 years ago
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