Why do we use microwaves to communicate berween earth and satellites
2 answers:
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Answer:
The spring's maximum compression will be 2.0 cm
Explanation:
There are two energies in this problem, kinetic energy and elastic potential energy
(with m the mass, v the velocity, x the compression and k the spring constant. ) so the total mechanical energy at every moment is the sum of the two energies:
Here we have a situation where the total mechanical energy of the system is conserved because there are no dissipative forces (there's no friction), so:
Note that at the initial moment where the hockey puck has not compressed the spring all the energy of the system is kinetic energy, but for a momentary stop all the energy of the system is potential elastic energy, so we have:
(1)
Due conservation of energy the equality (1) has to be maintained, so if we let k and m constant x has to increase the same as v to maintain the equality. Therefore, if we increase velocity to 2v we have to increase compression to 2x to conserve the equality. This is 2(1.0) = 2.0 cm
Answer:
0.00915 m
Explanation:
= Wavelength of light = 610 nm
L = Distance of slit to screen = 1.5 m
a = Slit gap = 0.2 mm
Width of central maximum is given by
The width of the central maximum is 0.00915 m
Answer:
a) h = 13,205.4 m
b) r_f = 2.12 106 m
c) e% = 0.68%
Explanation:
a) This is an exercise we are asked to use energy conservation,
Starting point. On the surface of Mimas
Em₀ = K = ½ m v²
Final point. Where the ball stops
= U = m g h
Em₀ = Em_{f}
½ m v² = m g h
h = ½ v² / g
let's calculate
h = ½ 41² / 0.0636
h = 13,205.4 m
b) For this part we are asked to use the law of universal gravitation, write the energy
starting point. Satellite surface
Em₀ = K + U = ½ m v² - GmM / r_o
final point. Where the ball stops
= U = - G mM / r_f
Em₀ = Em_{f}
½ m v² - G m M / r_o = - G mM / r_f
In this case all distances are measured from the center of the satellite
1 / rf = 1 / GM (-½ v² + G M / r_o)
let's calculate
1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)
1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)
1 / r_f = 4,714 10⁻⁷
r_f = 1 / 4,715 10⁻⁷
r_f = 2.12 106 m
to measure this distance from the satellite surface
r_f ’= r_f - r_o
r_f ’= 2.12 106 - 1.98 105
r_f ’= 1,922 106 m
c) the percentage difference is
e% = 13 205.4 / 1,922 106 100
e% = 0.68%
The estimate of part a is a little low