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attashe74 [19]
2 years ago
10

A girl is bouncing on a trampoline where is her gravitational potential energy a maximum and where is her kinetic energy maximum

Physics
1 answer:
Stella [2.4K]2 years ago
7 0

Answer:

When you jump down, your kinetic is converted to potential energy of the stretched trampoline. The trampoline's potential energy is converted into kinetic energy, which is transferred to you, making you bounce up. At the top of your jump, all your kinetic energy has been converted into potential energy. Right before you hit the trampoline, all of your potential energy has  been converted back into kinetic energy. As you jump up and down your kinetic energy increases and decrease.

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A parachute works because the canvas of the parachute is acted upon by __________.
Ne4ueva [31]
The question is asking to choose among the following choices that could complete the question about the inertia, base on my research and further investigation, the possible answer would be letter B. Gravity. I hope you are satisfied with my answer and feel free to ask for more 
8 0
2 years ago
Read 2 more answers
An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init
elena-14-01-66 [18.8K]

Answer:

Approximately 71\%.

Explanation:

The formula for the kinetic energy \rm KE of an object is:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2,

where

  • m is the mass of that object, and
  • v is the speed of that object.

Important: Joule (\rm J) is the standard unit for energy. The formula for \rm KE requires two inputs: mass and speed. The standard unit of mass is \rm kg while the standard unit for speed is \rm m \cdot s^{-1}. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

m = 63\; \rm g = 0.063\; \rm kg.

Initial \rm KE of this arrow:

\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}.

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%.

8 0
2 years ago
A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.610 c. A scientist on Coruscant measures th
mamaluj [8]

Answer:

the length of the now stationary spacecraft = 89.65m

Explanation:

In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.

Thus, it has a formula;

L = L_o(√(1 - (v²/c²))

Where in this question;

L = 71m and v = 0.610 c

Thus;

71 = L_o (√(1 - ((0.61c)²/c²))

c² will cancel out to give;

71 = L_o (√(1 - 0.61²)

71 = L_o (√(1 - 0.61²)

71 = 0.792L_o

L_o = 71/0.792

L_o = 89.65m

6 0
2 years ago
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on it
Nadya [2.5K]

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

4 0
2 years ago
The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length
Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

y_r=1.22\frac{\lambda f}{d}

Where,

y_r = Range of the radius

\lambda = wavelength

f= focal length

Our values are given by,

State 1:

d=2.00mm = 2*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 750nm = 750*10^{-9}m

State 2:

d=8.00mm = 8*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 390nm = 390*10^{-9}

Replacing in the first equation we have:

y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}

y_{r1}= 11.4\mu m

And also for,

y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}

y_{r2} = 1.49\mu m

Therefor, the airy disk radius ranges from 1.49\mu m to 11.4\mu m

7 0
2 years ago
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