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Lelu [443]
2 years ago
11

A kite is flying 79 ft off the ground, and its string is pulled taut. The angle of elevation of the kite is 50°. Find the length

of the string. Round your answer to the nearest tenth.
Mathematics
1 answer:
Art [367]2 years ago
4 0

Answer:

If the kite has an 11 ft line attached to it and it casts an 8 feet shadow, we will have a rectangle triangle (see picture below).

To find the height x of the kite, we can apply the Pythagorean theorem.

x² = 11² - 8²

x² = 121 - 64

x² = 57

x =√57

Step-by-step explanation:

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Answer: (a,a)


Step-by-step explanation:

Given: A  right isosceles triangle which is placed on the coordinate plane such that one of the legs of the triangle lies along the x-axis and the other leg is parallel to the y-axis.

From the given figure one vertex is on origin(0,0) and second on x -axis as (a,0) .ii.e. one side of the triangle is on the x axis with length 'a' and the side parallel to the y axis should perpendicular to the x axis with length 'a'.(∵ two sides are equal in a right isosceles triangle)

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3 years ago
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A 0. 0427 kg racquetball is moving 22. 3 m/s when it strikes a stationary box. The ball bounces back at 11. 5 m/s, while the box
andreyandreev [35.5K]

To solve the question we must know about the conservation of momentum.

<h2>Conservation of Momentum</h2>

According to the law of conservation of momentum, when two bodies collide with each other then their momentum is conserved if no external force acts on the system. therefore, the combined momentum of the two objects before collision will be equal to the combined momentum of the two objects after the collision.

the momentum of the two objects before the collision

                       = momentum of the two objects after the collision

The mass of the box is 0.30141 kg.

<h2>Explanation</h2>

Given to us

  • Mass of racquetball, m = 0.0427 kg,
  • Velocity of racquetball before Collision,  \bold{u_1} = 22.3 m/s,
  • Velocity of box before Collision,\bold{ u_2}  = 0 m/s,
  • Velocity of racquetball after Collision, \bold{v_1} = 11.5 m/s,
  • Velocity of box after Collision, \bold{v_2} = 1.53 m/s,

<h3>Assumption</h3>

Let the mass of the box be M.

<h3>Conservation Of Momentum</h3>

Momentum before collision = Momentum after collision

mu_1 + Mu_2 = mv_1 + Mv_2

given, the box was stationary before the collision,

mu_1 + Mu_2 = mv_1 + Mv_2\\\\&#10;mu_1 + 0 = mv_1 + Mv_2\\\\&#10;mu_1  - mv_1 = Mv_2\\\\&#10;M= \dfrac{mu_1  - mv_1}{v_2}

Substituting the values,

M= \dfrac{mu_1  - mv_1}{v_2}\\\\&#10;M = \dfrac{(0.0427 \times 22.3)  - (0.0427\times 11.5)}{1.53}\\\\&#10;M = 0.30141\ kg

Hence, the mass of the box is 0.30141 kg.

Learn more about Conservation of Momentum:

brainly.com/question/2141713

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Step-by-step explanation:

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