Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
Answer:
e*P_s = 11 W
Explanation:
Given:
- e*P = 1.0 KW
- r_s = 9.5*r_e
- e is the efficiency of the panels
Find:
What power would the solar cell produce if the spacecraft were in orbit around Saturn
Solution:
- We use the relation between the intensity I and distance of light:
I_1 / I_2 = ( r_2 / r_1 ) ^2
- The intensity of sun light at Saturn's orbit can be expressed as:
I_s = I_e * ( r_e / r_s ) ^2
I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2
I_s = 11 W / e*a
- We know that P = I*a, hence we have:
P_s = I_s*a
P_s = 11 W / e
Hence, e*P_s = 11 W
Answer:
the required mass flow rate is 49484.37 kg/s
Explanation:
Given the data in the question;
we first determine the relation for mass flow rate of water that passes through the turbine;
so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;
= m ( Δ P.E )
so we substitute (gh) for ( Δ P.E );
= m (gh)
m = / gh
so we substitute our given values into the equation
m = 100 MW / ( 9.81 m/s²) × 206 m
m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m
m = 10 × 10⁷ / 2020.86
m = 49484.37 kg/s
Therefore, the required mass flow rate is 49484.37 kg/s
Answer:
The radius of the curve that Car 2 travels on is 380 meters.
Explanation:
Speed of car 1, 
Radius of the circular arc, 
Car 2 has twice the speed of Car 1, 
We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :
According to given condition,
On solving we get :
So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.
<span>Answer: 1600 J
Explanation:
1) Data:
a) ideal gas: ⇒ pV = nRT and work = ∫ pdV
b) slowly compressed ⇒ constant temperature and not heat exchange
c) pressure: p = 2 atm
d) intitial volume: Vi = 10 liters
e) final volumen: Vf = 2 liters.
f) then the
volume of the gas is held constant ⇒ not work in this stage.
g) calculate the work done on the gas: W = ?
2) Equation
W = ∫pdV
3) Solution:
Since p = constant, W = p ∫dV = p ΔV = p (Vf - Vi)
p = 2 atm × 1.0 ×10⁵ Pa / atm = 200.000 Pa
Vi = 10 liter × 0.001 m³ ./ liter = 0.01 m³
Vf = 2 liter × 0.001 m³ / liter = 0.002 m³
W = 200.000 Pa × (0.002 m³ - 0.01m³) = - 1.600 J.
The negative sign means the work is done over the system.
That is all the work in the system because at the second stage the volume is held constant.
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