Answer:
15.3 m/s
Explanation:
Radius of orbit= 6400+6300 = 12700 km
Circumference of orbit= 2*(22/7)*12700 =79796.45*10^3 m
Now,
Speed= Distance / Time
= 79796.45*10^3/(24*60*3600)
= 15.3 m/s
Answer:
yes
Explanation: Work is done when there is movement. Therefore it was work was being done.
Answer:
Aphelion: 6404 W/m2
Perihelion: 14978 W/m2
Explanation:
The solar energy flux depends on the solar power output divided by the surface of a sphere with a radius equal to the distance to the Sun.
The distances we need are the aphelion and perihelion of Mercury.
Planetary orbits are ellipses. In an ellipse the eccentricity is related to linear eccentricity and the length of the semi major axis:
Where
e: eccentricity
c: linear eccentricity
a: semi major axis
The linear eccentricity is equal to the distance of the focus of the center of the ellipse.
a = 0.39 AU = 5.83e10 m
In planetary orbits the Sun is in one of the fucuses. With this we can calculate the prihelion and aphelion as:
Ap = a + c = 5.83e10 + 1.22e10 = 7.05e10 m
Pe = a - c = 5.83e10 - 1.22e10 = 4.61e10 m
And the solar energy fluxes will be:
Hi There,
This is False.
Hope this helped!
Answer:
a) dh/dt = -44.56*10⁻⁴ cm/s
b) dr/dt = -17.82*10⁻⁴ cm/s
Explanation:
Given:
Q = dV/dt = -35 cm³/s
R = 1.00 m
H = 2.50 m
if h = 125 cm
a) dh/dt = ?
b) dr/dt = ?
We know that
V = π*r²*h/3
and
tan ∅ = H/R = 2.5m / 1m = 2.5 ⇒ h/r = 2.5
⇒ h = (5/2)*r
⇒ r = (2/5)*h
If we apply
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = 3*35/π = 105/π ⇒ d(r²*h)/dt = -105/π
a) if r = (2/5)*h
⇒ d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π
⇒ (4/25)(3*h²)(dh/dt) = -105/π
⇒ dh/dt = -875/(4π*h²)
b) if h = (5/2)*r
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π
⇒ (5/2)*(3*r²)(dr/dt) = -105/π
⇒ dr/dt = -14/(π*r²)
Now, using h = 125 cm
dh/dt = -875/(4π*h²) = -875/(4π*(125)²)
⇒ dh/dt = -44.56*10⁻⁴ cm/s
then
h = 125 cm ⇒ r = (2/5)*h = (2/5)*(125 cm)
⇒ r = 50 cm
⇒ dr/dt = -14/(π*r²) = - 14/(π*(50)²)
⇒ dr/dt = -17.82*10⁻⁴ cm/s
