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melamori03 [73]
3 years ago
14

A. A man of 100 kg mass is standing on the surface of the earth of mass 6 × 1024 kg and

Physics
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

m1=100kg

m2=6×10^24kg

r=6400km

G=6.6×10^-11

F=?

as we know that

F=G m1.m2/r²

F=6.6×10^-11×100×6×10^24/(6400)²

F=6.6×10^-11×6×10^26/4.096×10^7

F=6.6×10^-11×1.465×10^19

F=9.67×10^8N

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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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If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?
Lady_Fox [76]
·The acceleration of gravity is proportional to

                       1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is  9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

                                                             =      490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

                 (9.8 m/s²) · (1/5)²  =  0.39 m/s² .

The boy's weight is    (mass) · (gravity)  =  (50kg) · (0.39 m/s²)

                                                                  =     19.6 newtons .

Just as we expected, his weight at that distance is

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A bomb of mass 4 kg, initially at rest, explodes breaking into two fragments of 1 kg and 3 kg. Which one of the following statem
Kazeer [188]

Answer: The 1 kg fragment will have three times the speed of the 3kg fragment.

Explanation:Here for the bomb, its chemical energy gets converted into the mechanical energy.

According to the law of conservation of momentum, the two bodies will have equal momentum and to satisfy this condition the lighter mass will have the higher velocity.

∵ momentum, p = mass × velocity

∴The 1 kg fragment will have three times the speed of the 3kg fragment.

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