The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a
1 answer:
Answer:
x_{cm} = 4.644 10⁶ m
Explanation:
The center of mass is given by the equation
= 1 /
∑
Where M_{total} is the total masses of the system, is the distance between the particles and
is the masses of each body
Let's apply this equation to our problem
M = Me + m
M = 5.98 10²⁴ + 7.36 10²²
M = 605.36 10²² kg
Let's locate a reference system located in the center of the Earth
Let's calculate
x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]
x_{cm} = 4.644 10⁶ m
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1) Maximum
2) Maximum
Explanation:
The force acting on a mass on a spring is given by Hooke's law; in magnitude:
where
F is the force
k is the spring constant
x is the displacement
Also we know from Newton's second law that we can write
where
m is the mass
a is the acceleration
So we can write the equation as
(1)
From this relationship, we see that the acceleration is directly proportional to the displacement.
On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by
(2)
where the first term is the elastic potential energy while the second term is the kinetic energy, and where
v is the velocity of the mass
From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.
Therefore this means that:
- When acceleration increases, velocity decreases
- When acceleration decreases, velocity increases
Therefore, the two answers here are:
- When the acceleration of a mass on a spring is zero, the velocity is at a maximum
When the velocity of a mass on a spring is zero, the acceleration is at a maximum
Answer:
The initial velocity vector of the ball is;
= 25·i + 19.6·j
Explanation:
The given parameters are;
The time of flight of the ball = 4 seconds
The horizontal distance from the plate at which the ball is caught = 100 m
Let 'u', represent the initial velocity, we have;
u × cos(θ) × t = u × cos(θ) × 4 s = 100 m
u × cos(θ) = 25 m/s...(1)
∴ 2·u·sin(θ) = 9.8 m/s² × 4 s = 39.2 m/s
tan(θ) = 1.568/2 = 0.784
θ = arctan(0.784) ≈ 38.096°
The direction of the velocity vector of the ball, θ ≈ 38.096°
From equation (1), we have;
u × cos(θ) = 25 m/s
∴ u = 25 m/s/cos(θ) = 25 m/s/cos(arctan(0.784)) = 31.7672787629 m/s
The magnitude of the initial velocity vector, u = 31.7672787629 m/s
The vertical component of the initial velocity, = u × sin(θ)
∴ = 31.7672787629 × sin(arctan(0.784))
Therefore, the initial velocity vector of the ball is approximately, v = 31.767 m/s in a direction 38.096° above the horizontal, from which we have;
u = uₓ·i + ·j = 25·i + 19.6·j
= 25·i + 19.6·j