What will be the acceleration of a body moving with uniform velocity?
1 answer:
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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
Answer
given,
wavelength = λ = 18.7 cm
= 0.187 m
amplitude , A = 2.34 cm
v = 0.38 m/s
A) angular frequency = ?
angular frequency ,
ω = 2π f
ω = 2π x 2.03
ω = 12.75 rad/s
B) the wave number ,
C)
as the wave is propagating in -x direction, the sign is positive between x and t
y ( x ,t) = A sin(k x - ω t)
y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)
Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer,
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :
Time period of oscillation measured by the observer is :
So, the time period of oscillation measured by the observer is 5.79 seconds.