<span>f2 = f0/4
The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius.
The expression for the force of gravity is
F = G*m1*m2/r^2
where
F = Force
G = Gravitational constant
m1,m2 = masses involved
r = distance between center of masses.
Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after.
f0 = G*m1*m2/r^2
f2 = G*m1*m2/(2r)^2
f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2)
The Gm m1, and m2 terms cancel, so
f2/f0 = (1/(2r)^2) / (1/r^2)
f2/f0 = (1/4r^2) / (1/r^2)
And the r^2 terms cancel, so
f2/f0 = (1/4) / (1/1)
f2/f0 = (1/4) / 1
f2/f0 = 1/4
f2 = f0*1/4
f2 = f0/4
So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
Answer:
Explanation:
A )
The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³
B )
Tension in the ball will be equal to net force acting on the ball
Net force on the ball = buoyant force - weight .
4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )
= 40.65 x 10⁻⁶ N .
C )Tension in the 3 rd ball will be equal to net force acting on the ball
Net force on the ball = weight - buoyant force
= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000 )
= 121.6 x 10⁻⁶ N .
Answer:
1. 579 x 10 ^-22N
Explanation:
F = kq1q2/r^2
= 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2
= 1. 579 x 10 ^-22N
Answer:
r= 2.17 m
Explanation:
Conceptual Analysis:
The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:
E= 2k*(λ/r) Formula (1)
Where:
E: electric field .( N/C)
k: Coulomb electric constant. (N*m²/C²)
λ: linear charge density. (C/m)
r : distance from the charge line to the surface where E calculates (m)
Known data
E= 2.9 N/C
λ = 3.5*10⁻¹⁰ C/m
k= 8.99 *10⁹ N*m²/C²
Problem development
We replace data in the formula (1):
E= 2*k*(λ/r)
2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)
r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)
r= 2.17 m
Efficiency η of a Carnot engine is defined to be:
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>
<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>
<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>
<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
