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2 years ago

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

Physics

1 answer:

zheka24 [161]2 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

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Explain Rutherford's experiment?

Ipatiy [6.2K]

Answer:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

Explanation:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.

The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.

Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.

7 0

2 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago

Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of

Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

$PV=nRT$

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

$V = 21\% * 0.66L$

$V = 0.1386L$

Our values are given as

P = 1atm

T=293K $R = 0.083145kJ*mol^{-1}K^{-1}$

Using the equation to find n, we have:

$PV=nRT$

$n = \frac{PV}{RT}$

$n = \frac{(1)(0.1386)}{(0.0821)(293)}$

$n = 5.761*10^{-3}mol$

Number of molecules would be found through Avogadro number, then

$\#Molecules = 5.761*10^{-3}*6.022*10^{23}$

$\#Molecules = 3.469*10^{21} molecules$

7 0

2 years ago

The teachers of Scott, Christopher, Dianne, and Kailee have last names Koeninger, Dannemiller, Briscoe, and Carter. Match the ki

Pani-rosa [81]

Scott-Dannemiller, Koeninger, Briscoe, and Carter

Christopher-Briscoe, Dannemiller, and Koeninger

Dianne-Koeninger, Briscoe, and Carter

Kailee-Koeninger and Carter

4 0

3 years ago

four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?

Kay [80]

There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.

3 0

3 years ago

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