Answer:
El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Explanation:
El gasto es el flujo volumétrico de gasolina (
), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:
(1)
Donde:
- Diámetro de la manguera, medido en pies.
- Velocidad medida de salida, medida en pies por segundo.
Si sabemos que
y
, entonces el gasto de gasolina es:


El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Answer:
3540.5N
Explanation:
Step one:
given data
mass m= 0.196kg
speed v= 31m/s
distance r= 5.32cm = 0.0532m
Step two
The expression relating force, mass, velocity and distance is
F= mv^2/r
substitute we have
F=0.196*31^2/0.0532
F=0.196*961/0.0532
F=188.356/0.0532
F=3540.5N
Answer:
d = 10.2 m
Explanation:
When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

where,
m = mass of car
v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)
v = 10 m/s
F = force on the car in direction of inclination = W Sin θ
W = weight of car = mg
θ = Angle of inclinition = 30°
d = distance covered up the ramp = ?
Therefore,

<u>d = 10.2 m</u>
Answer:
8.00 kJ
Explanation:
The first thing is to determine what quantities are there.
the mass of water = 1 000 kg
initial velocity, u = 6 m/s
final velocity, v = 4 m/s
the generator is operating at 100 % efficiency, so there is no energy loss.
The kinetic energy, Ek is converted to electrical energy, therefore Ek = electrical energy.
The kinetic energy is calculated as follows:
Ek = 1/2 mv²
= 1/2×(1 000)× (4)²
= 8 000 J/s
= 8.00 kJ Ans
Answer:
W = 8.92 10² kJ
Explanation:
For this exercise they give us the strength, we must calculate the distance traveled, for this we need the rocket acceleration let's use Newton's second law
F = m a
a = F / m
a = 20 103/1400
a = 14.29 m/s²
With kinematics we can find the distance traveled
² = v₀² + 2 a x
x = (
²-v₀²) / 2 a
x = (50² -35²) / 2 14.29
x = 1275 / 28.58
x = 44.61 m
Let's calculate the work
W = F.d
The bold is vector; as indicated by the force is in the direction of movement the scalar product is reduced to the ordinary product
W = F d
W = 20 10³ 44.61
W = 8.92 10⁵ J
W = 8.92 10² kJ