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LuckyWell [14K]
2 years ago
11

A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the

mass of its planet from this information.
Physics
1 answer:
kap26 [50]2 years ago
5 0

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

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Which situation would create a field like the one shown here? A. an interaction between two magnetic south poles B. an interacti
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A 60.0-kg mass person wishes to push a 120-kg mass box across a level floor. The coefficient of static friction between the pers
Aneli [31]

Answer:

μ = 0.350

Explanation:

For the person to able to move the box, the force exerted by the person on the box must equal the force exerted by the box:

F_{p} = F_{b}

In this case, force can be calculated as a product of mass (m) by the acceleration of gravity (g) and the coefficient of static friction (μ):

m_{p}*g*\mu_{p}=m_{b}*g*\mu_{b}\\m_{p}*\mu_{p}=m_{b}*\mu_{b}\\60*0.7=120*\mu_{b}\\\mu_{b}= 0.35

Therefore, for the person to be able to push the box horizontally, the coefficient of static friction between the box and the floor should not be higher than 0.350.

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2 years ago
A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r
Firlakuza [10]

Answer:

Speed of the this part is given as

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

P_1 + P_2 + P_3 = P_i

here we know that

P_1 = P_2

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

P_1 + P_2 = \sqrt{P^2 + P^2}

P_1 + P_2 = \sqrt2 mv

now from above equation we have

P_3 = -(P_1 + P_2)

mv_3 = -(\sqrt 2 mv)

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

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2 years ago
The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe
Novosadov [1.4K]

Answer:

a) a= 8.33 m/s²,    T = 12.495 N , b)    a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

            fr₁ + fr₂ = ma

The equation for the force of friction is

          fr = μ N

Y Axis

         N - W₁ –W₂ = 0

         N = W₁ + W₂

         N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

        N = 2 m g

We replace

           μ₁ 2mg + μ₂ mg = m a

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          T = m₂ a

          T = 1500 8.33

          T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

           fr = m₂ a₂

           N- w₂ = 0

          N = W₂ = mg

          μ₂ mg = m a₂

          a = μ₂ g

          a = 0.25 9.8

          a = 2.45 m / s²

4 0
2 years ago
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