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Firdavs [7]
3 years ago
14

Two satellites are in circular orbits around the earth. the orbit for satellite a is at a height of 540. km above the earth's su

rface, while that for satellite b is at a height of 890. km. find the orbital speed for satellite a and satellite
b.
Physics
1 answer:
kotegsom [21]3 years ago
7 0
<span>Answer: Va = 7,625 m/s Vb = 7,404 m/s Given: A = 486,000 m B = 901,000 m G = 6.67428E-11 m^3/kg-s^2 M = 5.9736E+24 kg r = 6,371,000 m Recall that you need the actual orbital distance from the *center* of the Earth, giving radius plus altitude: rA = 6,857,000 m rB = 7,272,000 m Equation: V = SQRT { GM / r } Solve for A Va = SQRT { [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (6,857,000 m) } Va = SQRT { [ 3.9869 m^3/s^2 ] / (6,857,000 m) } Va = SQRT { 58,144,202 m^2/s^2 } Va = 7,625 m/s Solve for B Vb = SQRT { [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (7,272,000 m) } Vb = SQRT { [ 3.9869 m^3/s^2 ] / (7,272,000 m) } Vb = SQRT { 54,826,016 m^2/s^2 } Vb = 7,404 m/s</span>
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Find the cp of an article which is sold for rupees 4048 , thereby , making a profit of 15 percent​
mestny [16]

Answer:

4048 = 115x/100.

Explanation:

Profit P = (15/100) X x.

P= 15x/100.

15x/100 = 4048- x.

4048 = 15x/100 + x.

4048 = 115x/100.

6 0
3 years ago
The body weighing 2 kg moves through the horizontal surface and crosses the path x = 75 cm The coefficient of friction of the bo
Taya2010 [7]

The kinetic energy of the body in definitive position is 4.24 J.

Explanation:

As per the work energy theorem, the work done on any system or object to move it from one position to another is equal to the change in kinetic energy of the object. In this case, the body weighing 2 kg is moved over an horizontal surface for a distance of 75 cm. As there will be frictional force acting on the body while moving over the surface. This frictional force multiplied by the distance the object is moved will give the work done on the body.

Frictional force = Coeffficent of friction × Normal force.

As the weight of the body is 2 kg, the normal force acting on it will be mass multiplied with acceleration due to gravity.

Frictional force = - 0.8×9.8 × 2 =-15.68 N

So the work done will be the product of frictional force with the displacement of 75 cm or 0.75 m.

Work done =  Frictional force × Displacement

Work done = -15.68×0.75 = -11.76 J.

So the work is done by the object.

If the kinetic energy of the body at starting is 16 J, then the kinetic energy of the body at definitive position will be obtained as below.

Work done = change in kinetic energy

-11.76 J = Final kinetic energy-16 J

Final Kinetic energy = - 11.76+16

Final kinetic energy = 4.24 J

Thus, the kinetic energy of the body in definitive position is 4.24 J.

3 0
3 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
3 years ago
Two particles move along an x axis. The position of particle 1 is given by x ! 6.00t 2 # 3.00t # 2.00 (in meters and seconds); t
s344n2d4d5 [400]

Answer:

15.6m/s

Explanation:

V1=\frac{dx}{dt}=\frac{d}{dt}(6t^{2}+3t+2) because the derivate of the position is the velocity

V1=12t+3

V2=20+\int\limits^_ {} \,-8t because the integral of the acceleration is the velocity

V2=20-4t^{2}

V1=V2 to see when the velocities of particles match

12t+3=20-4t^2

4t^2+12t-17=0 we resolve this with \frac{-b+-(\sqrt{b^{2} -4ac} )}{2a}

and we take the positif root

t=1.05 sec

if we evaluate the velocity (V1 or V2) the result is 15.6m/s

7 0
3 years ago
What is a disadvantage associated with many renewable or alternative energy sources
Alex Ar [27]
Its pretty expensive and can't be replaced immediately and some source like nuclear energy produces nuclear waste which produce radioisotope that is harmful for all living beings
5 0
4 years ago
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