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astraxan [27]
3 years ago
13

Two satellites are in circular orbits around the earth. the orbit for satellite a is at a height of 542 km above the earth's sur

face, while that for satellite b is at a height of 838 km. find the orbital speed for satellite a and satellite
b.
Physics
1 answer:
Evgen [1.6K]3 years ago
6 0
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite 
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g' 
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h): 
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)] 
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec 
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
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Find the velocity in m/s of a swimmer who swims 110m toward the shore in 72s
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A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has
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The distance D where the object comes to rest is 1.08.m.

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(c) the distance D where the object comes to rest.

W_{total} =ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1*2.3^{2}

⇒D=\frac{0.5*2.3^{2} }{2.45}

⇒1.08.m

To learn more about distance, refer to:

brainly.com/question/4998732

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5 0
1 year ago
A small plastic bead has been charged to -15nC.
marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

(C) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.  

8 0
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