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ioda
3 years ago
8

Why is a human powered generator better than the sun (solar cells)?

Physics
1 answer:
zheka24 [161]3 years ago
6 0

Answer:

we can not use the suns energy too effectively in power cells and with human power we can generate more energy

You might be interested in
How much water can be held by a cylindrical tank with a radius of 12 feet and a height of 30 feet? A. 2,260.8 cubic feet B. 13,5
MakcuM [25]

Answer:

Volume, V = 13564.8  cubic feet

Explanation:

It is given that,

Radius of the cylindrical tank, r = 12 feet

Height of the tank, h = 30 feet

We need to find the water that can be held by a cylindrical tank i.e. we need to find the volume of the tank. It is given by :

V=\pi\times r^2\times h

V=3.14\times (12)^2\times 30

V = 13564.8  cubic feet

So, the water held by the tank is 13564.8  cubic feet. Hence, this is the required solution.

6 0
3 years ago
Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c
stira [4]

Answer:

i) 24.5 m/s

ii) 30,656 m

iii) 89,344 m

Explanation:

Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer

i) Los parámetros dados son;

Altura inicial, s = 120 m

El tiempo en caída libre = 2.5 s

De la ecuación de caída libre, tenemos;

v = u + gt

Dónde:

u = Velocidad inicial = 0 m / s

g = Aceleración debida a la gravedad = 9.81 m / s²

t = Tiempo de caída libre = 2.5 s

Por lo tanto;

v = 0 + 9.8 × 2.5 = 24.5 m / s

ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación

s = u · t + 1/2 · g · t²

= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m

iii) La altura restante = 120 - 30.656 = 89.344 m.

6 0
3 years ago
4. A car accelerating at 60 meters/second is hit in an accident by a bus. The net force exerted on
Licemer1 [7]

Answer:

500kg

Explanation:

mass = newtons/force divided by the acceleration rate

m = 30,000/60

m = 500

3 0
2 years ago
Which statements describe the motion of the object represented by the graph? Check all that apply.
Sav [38]

Answer:a, d, e

Explanation:

6 0
3 years ago
Read 2 more answers
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

8 0
3 years ago
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