When an object is in orbit, it is falling at the same rate at which the Earth is curving. Please select the best answer from the
1 answer:
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Answer:
g(h) = g ( 1 - 2(h/R) )
<em>*At first order on h/R*</em>
Explanation:
Hi!
We can derive this expression for distances h small compared to the earth's radius R.
In order to do this, we must expand the newton's law of universal gravitation around r=R
Remember that this law is:
In the present case m1 will be the mass of the earth.
Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):
Therefore, we can see that
With a the acceleration due to the earth's mass.
Now, the taylor series is going to be (at first order in h/R):
a(R) is actually the constant acceleration at sea level
and
Therefore:
Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:
The energy of the system is E=5.5 J. This energy is the same at every moment of the oscillation. When the stretch x of the spring is maximum (so, when the stretch is equal to the amplitude: x=A) the velocity of the spring is zero, so all this energy is just elastic potential energy of the spring:
From which we find
From which we find
Answer:
ΔT = 302 °c
Explanation:
mass (m) = 4.6 g = 0.0046 kg
velocity (v) = 278 m/s
specific heat of lead (c) = 128 J/kg. °c
kinetic energy = 0.5 mx
kinetic energy = 0.5 x 0.0046 x
kinetic energy = 177.8 J
since all the kinetic energy is converted to thermal energy,
kinetic energy = thermal energy (E) = 177.8 J
thermal energy = m x c x ΔT
where ΔT is the temperature change
177.8 = 0.0046 x 128 x ΔT
ΔT = 177.8 / 0.59
ΔT = 302 °c