To solve the problem, use Kepler's 3rd law :
T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
but first covert 6.00 years to seconds :
6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s
The radius of the orbit then is :
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m
Answer:
Explanation:
The angular speed is given by:
Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to 
:
Now, we calculate the angular speed:
Answer:
Explanation:
Since the sled plus passenger moves with constant velocity , force applied will be equal to frictional force. Let the force applied be F
a ) Frictional force = μ R = F cosφ
R = mg - F sinφ
μ(mg - F sinφ) = F cosφ
μmg = F (μsinφ+cosφ)
F = μmg / (μsinφ+cosφ)
Work done
= F cosφ x d
= μmg x cosφ x d / (μsinφ+cosφ)
b )Work done
= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / ( 0.13 sin36.7 +cos36.7)
= 1164.61 / .87946
1324.23 J
c ) work done on the sled by friction
= - (work done by force)
= - μmg x cosφ x d / (μsinφ+cosφ)
d ) work done on the sled by friction
= - 1324.23 J
Answer:
It is a predator.
Explanation:
They eat other insects so therefore I think they're a predator.
Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J
