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2 years ago

Plants and trees grow nearly everywhere. this is one of the advantages of what energy?

Physics

2 answers:

Naddik [55]2 years ago

6 0

Answer: Biomass

Explanation: Biomass is a plant or animal material that is used for the process of energy production. These source of energy production can be intentionally grown plants that can be used for the energy production.

Example: Miscanthus switchgrass.

Burning of the biomass into the environment generated carbon dioxide into the environment which is recycle again by the plants.

AnnyKZ [126]2 years ago

4 0

The answer is Biomass.

Letter :D

You might be interested in

I need to lift a 2000kg car, 1.798m and the joules required is 35240.8. Converted to watt (W = 35240.8/5 (s)) I got 7048.16 W. I

marusya05 [52]

This is a very interesting problem ... mainly because it's different from
the usual questions in the Physics neighborhood.

I can discuss it with you, but maybe not quite give you a final answer
with the information you've given in the question.

I agree with all of your calculations so far ... the total energy required,
and the power implied if the lift has to happen in 5 seconds.

First of all, let's talk about power. I'm assuming that your battery is
a "car" battery, and I'm guessing you measured the battery voltage
while the car was running. Turn off the car, and you're likely to read
something more like 13 to 13.8 volts.
But that's not important right now. What I'm looking for is the CURRENT
that your application would require, and then to look around and see whether
a car battery would be capable of delivering it.

   Power = (volts) x (current)

   7,050 W = (14 volts) x (current)

   Current = (7,050 watts / 14 volts) = 503 Amperes.

That kind of current knocks the wind out of me. I've never seen
that kind of number outside of a power distribution yard.
BUT ... I also know that the current demand from a car battery during
starting is enormous, so I'd better look around online and try to find out
what a car battery is actually capable of.

I picked a manufacturer's name that I'd heard of, then picked their
recommended battery for a monster 2003-model car, and looked at
the specs for the battery.

The spec I looked at was the 'CCA' ... cold cranking Amps.
That's the current the battery is guaranteed to deliver for 30 seconds,
at a temperature of 0°F, without dropping below 12 volts.

This battery that I saw is rated 803 Amps CCA !

OK. Let's back up a little bit. I'm pretty sure the battery you have
is a nominal "12-volt" battery. Let's say you use to start lifting the lift.
As the lift lifts, the battery voltage sags. What is the required current
if the battery immediately droops to 12V and stays there, while delivering
7,050 watts continuously ?

Power = (volts) x (current)

7,050 W = (12 V) x (current)

Current = (7,050 W / 12 V) = 588 Amps .

Amazingly, we may be in the ball park.
If the battery you have is rated by the manufacturer for 600 Amps
CCA (0°F) or CA (32°F), then the battery can deliver the current
you need.
BUT ... you can't conduct that kind of current through ear-bud wire,
or house wiring wire. I'm not even so sure of jumper-cables.
You need thick, no-nonsense cable, AND connections with a lot of
area ... No alligator clips. Shiny nuts and bolts with no crud on them.

Now ... I still want to check the matter of the total energy.
I'm sure you're OK, because the CCA and CA specifications talk about
30 seconds of cranking, and you're only talking about 5 seconds of lifting.
But I still want to see the total energy requirement compared to the typical
battery specification ... 'AH' ... ampere-hours.

You're talking about 35,000 joules

      = 35,000 watt-seconds

   = 35,000 volt-amp-seconds.

   (35,000 volt-amp-sec) x (1 hour/3600 sec) / (12 volt)

   = (35,000 x 1) / (3600 x 12) volt-amp-sec-hour / sec-volt

   =    0.81 Amp-Hour .

That's an absurdly small depletion from your car battery.
But just because it's only 810 mAh, don't get the idea that you can
do it with a few rechargeable AA batteries out of your camera.
You still need those 600 cranking amps. That would be a dead short
for a stack of camera batteries, and they would shrivel up and die.

Have I helped you at all ?

5 0

3 years ago

Will name brainliest and give 50 points

Natalija [7]

A I think it was sorry if not

3 0

3 years ago

Read 2 more answers

A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi

frosja888 [35]

Answer:

The answer to the question

The steady state response is u₂(t) = - $\frac{3\sqrt{2} }{2}$ cos(3t + π/4)

of the form R·cos(ωt−δ) with R = $-\frac{3\sqrt{2} }{2}$ , ω = 3 and δ = -π/4

Explanation:

To solve the question we note that the equation of motion is given by

m·u'' + γ·u' + k·u = F(t) where

m = mass = 2.00 kg

γ = Damping coefficient = 1

k = Spring constant = 3 N·m

F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)

Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by $\frac{-1+/-\sqrt{23} }{4} \sqrt{-1}$ = $\frac{-1+/-\sqrt{23} }{4} i$

This gives the general solution of the homogeneous equation as

u₁(t) = $e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)$

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which 2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t) we have

3·B-15·A = 27

3·A +15·B = 18

Solving the above linear system of equations we have

A = -1.5, B = 1.5 and u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = $-\frac{3\sqrt{2} }{2}$ ·cos(3·t + π/4)

The general solution is then u(t) = u₁(t) + u₂(t)

however since u₁(t) = $e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)$ ⇒ 0 as t → ∞ the steady state response = u₂(t) = $-\frac{3\sqrt{2} }{2}$ ·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = $-\frac{3\sqrt{2} }{2}$

ω = 3 and

δ = -π/4

8 0

3 years ago

What is a gravitational field and how its strength be measured

Yakvenalex [24]

A gravitational field is the field generated by a massive body, that extends into the entire space. Every object with mass m experiences a force F when immersed in a gravitational field. The intensity of the force is equal to
$F= \frac{GM}{r^2} m$
where $G=6.67 \cdot 10^{-11} m^3 Kg^{-1} s^{-2}$ is the gravitational constant, M is the mass of the source of the field (e.g. the mass of a planet), and r is the distance between the object and the source of the field. The force is always attractive.

A possible way to measure the intensity of a gravitational field is by measuring the acceleration a of the object immersed in this field. In fact, for Newton's second law we have:
$F=ma$
but since
$F= \frac{GM}{r^2} m$
we can write
$a = \frac{GM}{r^2}$
Therefore, by measuring the acceleration of the object, we also measure the intensity of the field.

5 0

3 years ago

A positive charge is moving across a room from south to

slega [8]

Answer:toward the ceiling

Explanation:

if you use the right hand rule it will point to the ceiling

8 0

2 years ago