A projectile is launched horizontally east at a speed of 29.4 M/s towards a wall 88.2 m away. What is the velocity of the projec
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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:
a) a = 5.03x10¹³ m/s²
b)
Explanation:
a) The acceleration of the positron can be found as follows:
(1)
Also,
(2)
By entering equation (1) into (2), we have:
<u>Where:</u>
F: is the electric force
m: is the particle's mass = 9.1x10⁻³¹ kg
q: is the charge of the positron = 1.6x10⁻¹⁹ C
E: is the electric field = 286 N/C
b) The positron's speed can be calculated using the following equation:
<u>Where</u>:
: is the final speed =?
: is the initial speed =0
t: is the time = 8.70x10⁻⁹ s
I hope it helps you!