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fgiga [73]
3 years ago
5

Imagine a girl climbing a rope ladder to ride a zip line. It takes effort for her to get to the top of the ladder, but then she

zips to the ground with ease. She travels down easily because she has stored energy. What’s your guess about where this energy came from?
Physics
2 answers:
Leokris [45]3 years ago
6 0

I think it comes from the stores energy from the gravity

Savatey [412]3 years ago
4 0

The stored energy is because of the girl’s elevated position. The energy was stored as she climbed to the top of the rope ladder.

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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m
Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5  *10⁻⁹C

r₁=0.840m

r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}

sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246

cos\beta =\frac{1}{\sqrt{1.64} } =0.7808

Calculation of the electric field at point P due to q1

Ep₁x=0

Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

Calculation of the electric field at point P due to q2

Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0
3 years ago
If you are sitting in a bus that is traveling along a straight, level road at 100 km/hr., you are traveling at 100 km/hr too. (a
aleksandr82 [10.1K]

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8 0
3 years ago
A. The potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J. Find by
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a. The spring is compressed  by  0.174 m.

b.  The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.

c. The horizontal velocity of the dart at that time is  13.74 m/s.

d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.

PE  of spring = 1/2 kx²

Put the values, we get

P.E = 0.940 = 1/2 x 62 x²

x = 0.174 m

Thus, the spring is compressed by 0.174 m.

b. Given is a 0.010 kg dart is fired straight up.

The vertical height is find out by

0.940 J = (0.010 kg) (9.8 m/s²) h

h = 9.6 m

Thus, the vertical distance the dart travels from its position is 9.6 m

c. From the conservation of energy principle, total mechanical energy is conserved.

1/2 mv² =mgh

v = √2gh

Plug the values, we get

v = √2 x 9.8x 9.6

v = 13.74 m/s

Thus, the horizontal velocity is  13.74 m/s.

d.  Time that dart spends in air, t = √2h/g

t = √(2x9.6)/9.81

t = 1.4 s

The horizontal distance from the equilibrium position at which the dart hits the ground.

Horizontal distance = (Velocity on x direction) x time

Horizontal distance = 13.74 m/s x 1.4s

Horizontal distance = 19.236 m

Thus, the horizontal distance is 19.236 m.

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Answer:

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An experiment is conducted such that an applied force is exerted on a 5kg object as it travels across a horizontal surface in wh
Katen [24]

If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.

The  graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.

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For more information on work done, visit

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2 years ago
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