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elixir [45]
3 years ago
15

Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

us 8.20×10^7 m, and orbital speed 4800 m/s. The second satellite has mass 54.0 kg and orbital radius 7.00×10^7 m.
A) What is the orbital speed of this second satellite?
v=?
Physics
1 answer:
Andreas93 [3]3 years ago
4 0

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite r_{1}= 8.20\times10^{7}

Orbital radius of second satellite r_{2}=7.00\times10^{7}\ m

Mass of first satellite m_{1}=53.0\ kg

Mass of second satellite m_{2}=54.0\ kg

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

v=\sqrt{\dfrac{GM}{r}}

From this relation,

v_{1}\propto\dfrac{1}{\sqrt{r}}

Now, \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}

v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}

Put the value into the formula

v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}

v_{2}=5195.16\ m/s

Hence, The orbital speed of this second satellite is 5195.16 m/s.

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The two conducting rails in the drawing are tilted upwards so they each make an angle of 30.0° with respect to the ground. The v
dolphi86 [110]

Answer:

The current flows through the rod is 14.9 A.

Explanation:

Given that,

Magnetic field = 0.045 T

Mass of aluminum rod  = 0.19 kg

Length = 1.6 m

Angle = 30.0°

We need to calculate the force

Using resolving force

F\cos\theta=mg\sin\theta

F=mg\tan\theta

Put the value into the formula

F=0.19\times9.8\times\tan30

F=1.075\ N

We need to calculate the current flows through the rod

Using formula of magnetic force

F=iLB

i=\dfrac{F}{LB}

Put the value into the formula

i=\dfrac{1.075}{1.6\times0.045}

i=14.9\ A

Hence, The current flows through the rod is 14.9 A.

8 0
3 years ago
A material that does not conduct electricity is called a... what?
xxTIMURxx [149]

A material that does not conduct electricity is known as an insulator

3 0
3 years ago
A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4
SIZIF [17.4K]

Answer:

The  value is    v = 47 \  m/s

Explanation:

From the question we are told that

   The initial  speed of the roller coaster is u =  13 \  m/s

    The  length of the hill is  l   = 400 \  m

    The  acceleration of the  roller coaster is a=4.0 \ m/s^2

Generally the acceleration is mathematically represented as

      a =  \frac{ v - u}{ t_f -  t_i }

Here  t_i is the initial time which is equal to zero

         v_f is the final velocity which is mathematically represented as

          v_f  =  \frac{d}{ t_f}

So  

     a =  \frac{ \frac{d}{d_f}  - u }{ t_f - t_i}

     4 = \frac{\frac{400}{ t_f}  - 13}{t_f - 0}

      4 =  \frac{400 - 13t_f}{ t_f} *  \frac{1}{t_f}

     4t_f ^2  +13f  + 400 =

Solving this using quadratic formula we obtain

    t_f =  8.5 \ s

     t_f =  -11.8 \ s

Generally  time cannot be negative so

       t_f =  8.5 \ s

Generally the  final velocity is mathematically represented as

         v = \frac{400}{8.5}

         v = 47 \  m/s

       

5 0
3 years ago
1) What are the parts of a longitudinal wave? How are they different?<br> ***
Kay [80]

Answer:

Compression- pushing together

Refraction- pulling apart

Wavelength- the length of the wave

Explanation:

I just knew it

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3 0
3 years ago
Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi
yanalaym [24]

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=v=6.1\times 10^7m/s

Charge on electron, q=e=1.6\times 10^{-19} C

Mass of electron,m_e=9.1\times 10^{-31} kg

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

V=\frac{v^2m_e}{2e}

Where V=Potential difference

m_e=Mass of electron

v=Velocity of electron

Using the formula

V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}

V=10581.59 V

Hence, the potential difference=10581.59 V

8 0
3 years ago
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