To determine the distance of the light that has traveled given the time it takes to travel that distance, we need a relation that would relate time with distance. In any case, it would be the speed of the motion or specifically the speed of light that is travelling which is given as 3x10^8 meters per second. So, we simply multiply the time to the speed. Before doing so, we need to remember that the units should be homogeneous. We do as follows:
distance = 3x10^8 m/s ( 8.3 min ) ( 60 s / 1 min ) = 1.494x10^11 m
Since we are asked for the distance to be in kilometers, we convert
distance = 1.494x10^11 m ( 1 km / 1000 m) = 149400000 km
Answer:
a) A = 4.0 m
, b) w = 3.0 rad / s
, c) f = 0.477 Hz
, d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1
²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
Answer:
v = 7.69 x 10³ m/s = 7690 m/s
T = 5500 s = 91.67 min = 1.53 h
Explanation:
In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:
where,
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Me = Mass of Earth = 5.97 x 10²⁴ kg
r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m
v = orbital speed = ?
Therefore,
<u>v = 7.69 x 10³ m/s</u>
For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.
So, its orbital speed can be given as:
where,
T = Time Period of Satellite = ?
Therefore,
<u>T = 5500 s = 91.67 min = 1.53 h</u>
Answer:
Magnetic field, B = 0.042 T
Explanation:
It is given that,
Speed of charged particle, 
Angle between velocity and the magnetic field, 
Charge, 
Magnetic force, F = 0.002 N
The magnetic force is given by :
B is the magnetic field
B = 0.042 T
So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.
Velocity, because if an object is in motion with no direction we will consider it as speed, but if it has direction we will consider it as Velocity. Hope it helps
