A 1.34 kg ball is connected by means of two massless strings, each of length L = 1.70m, to a vertical, rotating rod. The strings
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10,378.82 m/s is the second cosmic speed.
69,801 km is the radius of the synchronous orbit of a satellite.
Given
Mass of planet = 4.74 × kg
Radius of planet = 5870 km = 5870000m
First Cosmic speed = 7.34 km/sec
1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.
It can be calculated by
v = √2GM/r where,
v= Escape velocity of the satellite
G = Gravitational constant
M = Mass of planet
r = Radius of planet
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
2) Speed of the satellite at the given period
v = 2πr/T where,
T= Time period of rotation = 16.6 × 3600 seconds
r = v×T/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Hence
The <u>Second Cosmic Speed</u> i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.
And the radius of the synchronous orbit of a satellite is 69,801 km.
Learn more about cosmic speed here brainly.com/question/15351190
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Answer:
The answer is 1.87nm/s.
Explanation:
The water loss must be replaced by
of sap. 110g of sap corresponds to a volume of
thus rate of sap replacement is
The volume of sap in the vessel of length is
,
where is the cross sectional area of the vessel.
For 2000 such vessels, the volume is
taking the derivative of both sides we get:
on the left-hand-side is the velocity
of the sap, and on right-hand-side
; therefore,
and since the cross-sectional area is
;
therefore,
solving for we get:
which is the upward speed of the sap in each vessel.