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Lapatulllka [165]
2 years ago
14

Calculen si a nivel del mar se reciben del sol 1000 watts por metro cuadrado y un panel solar genera 200 watts, ¿Cuál es la dife

rencias.?
Physics
1 answer:
bogdanovich [222]2 years ago
7 0

Answer:

Diferencia de energía de las dos fuentes = 800 watts

Explanation:

Cantidad de energía recibida en el mar = 1000

Cantidad de energía solar recibida en el mar = 200

Diferencia de energía de las dos fuentes = 1000 -200 = 800 watts

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If an object is not moving, what’s the net force?
olasank [31]
No net force and the photo shows

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3 years ago
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A woman lifts a 300 newton child a distance of 1.5 meters in 0.75 seconds. What is her power output in lifting the child?
Nitella [24]
Power = Work done / time

Work done = Force * Distance
                 =  300 N *  1.5 m =  450 J

Power =  450 / 0.75 = 600 Watts.
7 0
3 years ago
. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
A child on a high dive has a mass of 40 kilograms. If the high dive is 10 meters in the air, what is the potential energy? GPE=m
saw5 [17]

Answer:

Ep = 3924 [J]

Explanation:

To calculate this value we must use the definition of potential energy which tells us that it is the product of mass by the acceleration of gravity by height.

E_{p}=m*g*h\\

where:

Ep = potential energy [J] (units of Joules)

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 10 [m]

E_{p} =40*9.81*10\\E_{p} = 3924 [J]

7 0
2 years ago
A driver must always stop within 50 ft but not less than ____________ ft from the nearest rail when the signal is flashing and t
k0ka [10]

Answer:

20 ft

Explanation:

5 0
3 years ago
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