Answer:
is the first confirmed terrestial planet to have been discovered outside the solar system by the kepler space telescope
Explanation:
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All three have the same current, so that is not a factor. Wattage (power) is E*I or i^2 R. The higher the resistance, the more power dissipated. The answer is R3 because it has the highest resistance.
R3 <<<< ===== answer.
Answer:
<em>The net force acting on the object is 0 N</em>
Explanation:
<u>Newton's Second Law of Forces</u>
The net force acting on a body is proportional to the mass of the object and its acceleration.
The net force can be calculated as the sum of all the force vectors in each rectangular coordinate separately.
The image shows a free body diagram where four forces are acting: two in the vertical direction and two in the horizontal direction.
Note the forces in the vertical direction have the same magnitude and opposite directions, thus the net force is zero in that direction.
Since we are given the acceleration a =0, the net force is also 0, thus the horizontal forces should be in equilibrium.
The applied force of Fapp=10 N is compensated by the friction force whose value is, necessarily Fr=10 N in the opposite direction.
The net force acting on the object is 0 N
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:
![F = G \frac{m_1m_2}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20G%20%5Cfrac%7Bm_1m_2%7D%7Br%5E2%7D)
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):
![F = m_2a](https://tex.z-dn.net/?f=F%20%3D%20m_2a)
Therefore, we can see that
![a(r) = G \frac{m_1}{r^2}](https://tex.z-dn.net/?f=a%28r%29%20%3D%20G%20%5Cfrac%7Bm_1%7D%7Br%5E2%7D)
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+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}](https://tex.z-dn.net/?f=a%28R%2Bh%29%20%5Capprox%20a%28R%29%20%2B%20h%20%5Cfrac%7Bda%28r%29%7D%7Bdr%7D_%7Br%3DR%7D)
a(R) is actually the constant acceleration at sea level
and
![a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}](https://tex.z-dn.net/?f=a%28R%29%20%3DG%20%5Cfrac%7Bm_1%7D%7BR%5E2%7D%20%5C%5C%20%5Cfrac%7Bda%28r%29%7D%7Bdr%7D_%7Br%3DR%7D%20%3D%20-2%20G%5Cfrac%7Bm_1%7D%7BR%5E3%7D)
Therefore:
![a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})](https://tex.z-dn.net/?f=a%28R%2Bh%29%20%5Capprox%20G%5Cfrac%7Bm_1%7D%7BR%5E2%7D%20-2G%5Cfrac%7Bm_1%7D%7BR%5E2%7D%20%5Cfrac%7Bh%7D%7BR%7D%20%3D%20g%281-2%5Cfrac%7Bh%7D%7BR%7D%29)
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:
![a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}](https://tex.z-dn.net/?f=a%28R%2Bh%29%20%5Capprox%20a%28R%29%20%2B%20h%20%5Cfrac%7Bda%28r%29%7D%7Bdr%7D_%7Br%3DR%7D%20%2B%20%5Cfrac%7Bh%5E2%7D%7B2%21%7D%20%5Cfrac%7Bd%5E2a%28r%29%7D%7Bdr%5E2%7D_%7Br%3DR%7D)