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Stolb23 [73]
3 years ago
13

Look at the picture below. The boy is swinging back and forth on the swing. At which point is the potential energy of a swing th

e greatest?
Point A (top)


Point B (halfway down)


Point C (bottom)


Point D (halfway up)
Physics
2 answers:
Komok [63]3 years ago
5 0
The correct answer is Point A the top
kvv77 [185]3 years ago
3 0

Point A (top)

Potential energy is the energy possessed by an object due to its height relative to a reference level. so greater the height above the reference level, greater will be the potential energy.

when the boy is at the top, the height is maximum there. hence the potential energy of the boy is maximum at the Top.

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VARVARA [1.3K]

Answer:

it's B. circuit a and b are series circuit while c is parallel

7 0
3 years ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

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3 years ago
Cliff divers at Acapulco jump into the sea macias (fjm793) – Homework 3, 2d motion 19-20 – dowd – (WoffordWPHY11920 2) 4 from a
Stells [14]

Answer:

v = 7.67 m/s

Explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

h = \frac{gt^2}{2}

solving for t

we have

t = \sqrt{\frac{2h}{g}}

substituing all value for time t

t = \sqrt{\frac{2\times 11.98}{9.8}}

t = 1.56 s

we know that speed is given as

v = \frac{d}{t}

v =\frac{11.98}{1.56}

v = 7.67 m/s

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Spiderman has a mass of 80 kg. He knows his webbing will break if it is exposed to a 200 N force. What is the maximum height he
likoan [24]

Answer:

This depends on the writers

if they want they can make spiderman deny the laws of nature

8 0
3 years ago
Convert planks constant in cgs system
dezoksy [38]

in cgs system, plank's constant= h=6.626 x10⁻²⁶ erg s

Value of Plank's constant in SI system= 6.626 x10⁻³⁴ Js

now 1 Joule= 10⁷ ergs

so h= 6.626 x10⁻³⁴ Js (10⁷ ergs/1J)

h=6.626 x10⁻²⁷ erg s

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