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MissTica
3 years ago
13

What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio

d of exactly 24.0 hours (86,400 seconds). (Measurement is from the center of Earth.) G = 6.67 × 10-11 N.m2/kg2 ME = 5.98 × 1024 kg 1 hr = 3,600 s
Physics
1 answer:
pantera1 [17]3 years ago
3 0
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),
 
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite. 

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite. 
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A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the
vitfil [10]

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}

v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}

\mathbf{v_{avg} =12.31 \ m/s}

8 0
3 years ago
To determine the types of equipment your boat must carry, you need to make a measurement of the boat. What is this measurement?
jenyasd209 [6]

Explanation:

In total, the length is measured from the tip of the bow in a linear fashion to the stern of the formation of delight including any back-deck extensions. The measurement involves bow sprits; rudders; detachable engines and engine sections; handles; and various fittings and connections.

Importance in calculating a boat's length:

it affects the transportation costs (the longer the length, the higher the cost).   The pontoon's length counts as you find out how much rope you need to wrestle.  

The cost of vessel settlement on marinas depends in part on the pontoon length. As more area is consumed by a more drawn pontoon, the docking charges are higher.  

Transportation guidelines will probably not allow pontoons past a specific length on specific occasions of the day.

6 0
3 years ago
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

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3 years ago
What type of stellar object has the greatest density
r-ruslan [8.4K]
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Fudgin [204]

D)

Explanation:

Coz the position changes with time but the position change is not constant throughout time So, D) does not have constant velocity

3 0
2 years ago
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