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charle [14.2K]
3 years ago
5

A planet has a mass of 5.68 x 1026 kg and a radius of 6.03 x 107 m. What is the weight of a 65.0 kg person on the surface of thi

s planet?
a.
636 N
c.
475 N
b.
678 N
d.
656 N
Physics
1 answer:
Paul [167]3 years ago
4 0

To solve this problem, we must know the gravitational force of the planet. The equation would be,

 F=G \frac{ m_{1} m_{2}  }{ r^{2} }

This would calculate the force between two objects with masses m1 and m2 and the gravitational constant, G, is 6.67 x 10^-11 m3 s-2 kg-1 and with r as the distance between the objects.

Thus,

F = (6.67 x 10^-11 m3 s-2 kg-1) * (5.68 x 10^26 kg) * (65 kg) * ((1/6.03 x 10^7 m)^2)

F = 678 kg/s^2 or 678 N

Answer is letter B.

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Answer:closed systems

Explanation:

A closed system is one in which matter does not enter or leave the system but there is exchange of energy between the system and its environment. In a closed system, the principle of energy conservation applies. The principle of energy conservation states that energy can neither be created nor destroyed but is converted from one form to another. An example of a closed system is a reaction vessel whose lid is closed.

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An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7
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Explanation:

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I=\frac{P_{avg}}{A}

I=\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}

I=7.296\times 10^{-6} W/m^2

Amplitude of electric field at receiver end

E_{max}=\sqrt{2I\mu _0c}

Amplitude of induced emf

=E_{max}d

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7 0
3 years ago
A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n
Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

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This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

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so we have

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so this is horizontal component of normal force is equal to the centripetal force on the car

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