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Wewaii [24]
3 years ago
5

Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface . How does the veloc

ity of Satellite A compare to the velocity of Satellite B ?
Physics
1 answer:
sp2606 [1]3 years ago
4 0

The velocity of Satellite A is 2% greater than velocity of satellite B.

The given parameters;

  • <em>Altitude of Satellite A = 500 km</em>
  • <em>Altitude of Satellite B = 800 km</em>

The forces acting on the Satellites are given as follows;

F_c = \frac{mv^2}{r} \\\\F_g = \frac{GMm}{r^2} \\\\\frac{v^2}{r}  = \frac{GM}{r^2}\\\\v^2 = \frac{GM}{r} \\\\v^2 r = GM\\\\v_1^2 r _1 = v_2^2 r_2\\\\v_A^2r_A = v_B^2 r_B\\\\(\frac{v_A}{v_B} )^2 = \frac{r_B}{r_A} \\\\\frac{v_A}{v_B}  = \sqrt{\frac{r_B}{r_A} } \\\\\frac{v_A}{v_B}  = \sqrt{\frac{(800,000\  + \ 6.4 \times 10^6}{(500,000\  + \ 6.4 \times 10^6} )} \\\\\frac{v_A}{v_B}  =  1.02 \\\\v_A = 1.02 \ v_B

v_A = v_B( 100\% \ + 2\%)\\\\v_A = 100\%v_B \ + \ \ 2\% v_B\\\\v_A = v_B \ \ + \ 2\% v_B

Thus, the velocity of Satellite A is 2% greater than velocity of satellite B.

Learn more about velocity of satellite here: brainly.com/question/13981089

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A bungee jumper attains a speed of 30 m/s just as the bungee cord begins to stretch. If the period of stretch is 2 s while comin
Bogdan [553]

Answer:1.53g

Explanation:

average deceleration= ?

inial velocity: u=0

final velocity: v=30m/s

time: t=2seconds

The first law of kinematics :

v=u+at

find a the subject of the formula

a=v-u/ t

a=\frac{0-30} 2

a=-30/2

a=-15m/s^{2}

The deceleration about g(acceleration due to gravity) will be:

15/9.8

1.53g

4 0
3 years ago
A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
galben [10]

Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

5 0
4 years ago
A 20.0 kg suitcase is raised 3.0 m above a platform. How much work is done on the suitcase?
Lisa [10]

Answer:

600

Explanation:

m=20

g=10m\s2

h=3m

we have

w=m*g*h

=20*10*3

=600joule

3 0
4 years ago
Including them, the mass of the balloon was 1890 kg and had a volume of 11,430 m3 . The balloon floats at a constant height of 6
Monica [59]

Given :

The mass of the balloon was 1890 kg and had a volume of 11,430 m3 .

The balloon floats at a constant height of 6.25m above the ground.

To Find :

The density of the hot air in the balloon.

Solution :

We know,

Volume × ( Density of surrounding air - Density of hot air ) = mass

Putting given values in above equation, we get :

11430\times ( 1.29 - \rho_{hot \ air } ) = 1890\\\\\rho_{hot \ air } = 1.29 - \dfrac{1890}{11430}\\\\\rho_{hot \ air } =  1.125\ kg\ m^3

Therefore, the density of hot air in the balloon is 1.125 kg m³.

4 0
3 years ago
Hi, i know that this isnt neces sary. but can anyone request a ga mes for me, im bo red. thankyou. ​
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Answer:

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5 0
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