Answer:
The kinetic energy is ![KE = 7.59 *10^{10} \ J](https://tex.z-dn.net/?f=KE%20%20%3D%20%207.59%20%20%2A10%5E%7B10%7D%20%5C%20%20J)
Explanation:
From the question we are told that
The radius of the orbit is ![r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m](https://tex.z-dn.net/?f=r%20%3D%20%202.3%20%2A10%5E%7B4%7D%20%5C%20km%20%20%3D%202.3%20%20%2A10%5E%7B7%7D%20%5C%20m)
The gravitational force is ![F_g = 6600 \ N](https://tex.z-dn.net/?f=F_g%20%20%3D%206600%20%5C%20N)
The kinetic energy of the satellite is mathematically represented as
![KE = \frac{1}{2} * mv^2](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20mv%5E2)
where v is the speed of the satellite which is mathematically represented as
![v = \sqrt{\frac{G M}{r^2} }](https://tex.z-dn.net/?f=v%20%20%3D%20%5Csqrt%7B%5Cfrac%7BG%20%20M%7D%7Br%5E2%7D%20%7D)
=> ![v^2 = \frac{GM }{r}](https://tex.z-dn.net/?f=v%5E2%20%20%3D%20%20%5Cfrac%7BGM%20%7D%7Br%7D)
substituting this into the equation
![KE = \frac{ 1}{2} *\frac{GMm}{r}](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A%5Cfrac%7BGMm%7D%7Br%7D)
Now the gravitational force of the planet is mathematically represented as
![F_g = \frac{GMm}{r^2}](https://tex.z-dn.net/?f=F_g%20%20%3D%20%5Cfrac%7BGMm%7D%7Br%5E2%7D)
Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
![KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A%5B%5Cfrac%7BGMm%7D%7Br%5E2%7D%5D%20%2A%20r)
=> ![KE = \frac{ 1}{2} *F_g * r](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2AF_g%20%2A%20r)
substituting values
![KE = \frac{ 1}{2} *6600 * 2.3*10^{7}](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A6600%20%2A%202.3%2A10%5E%7B7%7D)
![KE = 7.59 *10^{10} \ J](https://tex.z-dn.net/?f=KE%20%20%3D%20%207.59%20%20%2A10%5E%7B10%7D%20%5C%20%20J)
Answer:
![\theta = 25.3^\circ](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2025.3%5E%5Ccirc)
Explanation:
The acceleration of the block can be found by the kinematics equations:
![v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2](https://tex.z-dn.net/?f=v%20%3D%20v_0%20%2B%20at%5C%5C5.88%20%3D%201.26%20%2B%20a%281.1%29%5C%5Ca%20%3D%204.2~m%2Fs%5E2)
Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.
![F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ](https://tex.z-dn.net/?f=F%20%3D%20mg%5Csin%28%5Ctheta%29%20%3D%20ma%5C%5Cg%5Csin%28%5Ctheta%29%20%3D%20a%5C%5C%289.8%29%5Csin%28%5Ctheta%29%20%3D%204.2%5C%5C%5Ctheta%20%3D%2025.3%5E%5Ccirc)
Answer:
option (b) 4900 N
Explanation:
m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R
F = G Me x m / (R + h)^2
F = G Me x m / 2R^2
F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2
F = 4900 N
If the object is moving in a straight line at a constant speed, then that's
the definition of zero acceleration. It can only happen when the sum of
all forces (the 'net' force) on the object is zero.
And it doesn't matter what the object's mass is. That argument is true
for specks of dust, battleships, rocks, stars, rock-stars, planets, and
everything in between.
<span>To begin, the formula for finding frequency when wavelength is known is "f = c / w" when c is the constant velocity (3 * 10^8 m/s). To convert the wavelength into a common form (m/s), it will have to be multiplied by 10^-2. This leaves the equation as "f = 3.0 * 10^8 / (2.4 * 10^-5 * 10^-2), or 2.4 * 10^-7. This gives 1.25 * 10^15 m/s as the frequency.</span>