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Juliette [100K]

3 years ago

10

Photovoltaic energy is the conversion of _________ into electricity through a photovoltaic cell. A) sunlight B) hydropower C) wi

nd energy D) nuclear energy

1 answer:

Aleks04 [339]3 years ago

8 0

The answer is A) sunlight

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Effciency of a lever is never 100% or more. why?Give reason

Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below $100\%$ .

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let $D_\text{in}$ and $D_\text{out}$ denote these two distances, and let $F_\text{in}$ and $F_\text{out}$ denote the input and the output forces. If the lever is indeed idea, then:

$F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}$ .

Rearrange to obtain:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}$

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means $D_\text{in} > D_\text{out}$ , such that $F_\text{in} < F_\text{out}$ .

Despite $F_\text{in} < F_\text{out}$ , the amount of work required will stay the same. Let $s_\text{out}$ denote the required linear displacement for the output force. At a distance of $D_\text{out}$ from the fulcrum, the angular displacement of the output force would be $\displaystyle \frac{s_\text{out}}{D_\text{out}}$ . Let $s_\text{in}$ denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be $\displaystyle \frac{s_\text{in}}{D_\text{in}}$ . Because both the input and the output are on the same lever, their angular displacement should be the same:

$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

Rearrange to obtain:

$\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}$ .

While increasing $D_\text{in}$ reduce the size of the input force $F_\text{in}$ , doing so would also increase the linear distance of the input force $s_\text{in}$ . In other words, $F_\text{in}$ will have to move across a longer linear distance in order to move $F_\text{out}$ by the same $s_\text{out}$ .

The amount of work required depends on both the size of the force and the distance traveled. Let $W_\text{in}$ and $W_\text{out}$ denote the input and output work. For this ideal lever:

$\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}$ .

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be $100\%$ - not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, $F_\text{in}$ would be larger than ideal:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance})$ .

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below $100\%$ ,

4 0

3 years ago

The air around a pool and the water in the pool receive equal amounts of energy from the sun. Why does the air experience a grea

labwork [276]

Answer:

A

Explanation:

4 0

3 years ago

In major league baseball, the pitcher's mound is 60 feet from the batter.If a pitcher throws a 89 mph fastball, how much time el

lianna [129]

<span>Hitting can be broken down into three segments; SEE, REACT, SWING. You watch the ball in the pitcher's hand during the windup and you watch the ball leave the pitcher's hand when it is thrown and you watch the spin of the ball as it comes towards the plate. That is SEE. You determine what the pitch is (fastball, curveball, etc.), you determine where the ball is going to go and you determine whether it is headed towards an area that you think you can get good wood on it. That is REACT. You swing if you like the pitch or don't swing if you don't like the pitch. That is SWING. It doesn't make any difference whether you are playing with a tennis ball or a golf ball or a baseball. The theory and mechanics are the same. So, to answer your question I would say the more you practice the better you will be, regardless of the type of ball you use to practice with.</span>

8 0

3 years ago

I need to talk to a girl my age

Greeley [361]

Answer:

that is my question

Explanation:

you want a girl to talk your age

<h2>HOW OLD ARE YOU</h2>

6 0

3 years ago

When a garden hose with an output diameter of 20 mm is directed straight upward, the stream of water rises to a height of 0.13m

SpyIntel [72]

Answer: h = 0.52m

Explanation:

Using the equation of out flow;

A1 × V1 = A2 ×V2

Where A1 = area of the first nozzle

A2 = area of the second nozzle

V1= velocity of flow out from the first nozzle

V2 = velocity of flow out from 2nd nozzle

But AV= area of nozzle × velocity of water = volume of water per second(m³/s).

Now we can set A×V = Area of nozzle × height of rise.

Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)

D1 = 20mm= 0.02m; h1 = 0.13m

D2 = 10mm = 0.01m; h2= ?

h2 = π(D1/2)²× h1 /π(D2/2)²

h2 = (0.02/2)² × 0.13/(0.01/2)²

= (0.01)² ×0.13 /(0.005)²

= 1.3 × 10^-5/(5 × 10^-3)²

= 1.3 × 10^-5/25 × 10^-6

= (1.3/25) 10^-5 × 10^6

= 0.052× 10

= 0.52m

7 0

3 years ago