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3 years ago

Problems with solar energy include _____.

2 answers:

german3 years ago

4 0

First choice: the inability of current technology to capture
large amounts of the Sun's energy

Well, it's true that large amounts of it get away ... our 'efficiency' at capturing it is still rather low. But the amount of free energy we're able to capture is still huge and significant, so this isn't really a major problem.

Second choice: the inability of current technology to store
captured solar energy

No. We're pretty good at building batteries to store small amounts, or raising water to store large amounts. Storage could be better and cheaper than it is, but we can store huge amounts of captured solar energy right now, so this isn't a major problem either.

Third choice: inconsistencies in the availability of the resource

I think this is it. If we come to depend on solar energy, then we're
expectedly out of luck at night, and we may unexpectedly be out
of luck during long periods of overcast skies.

Fourth choice: lack of demand for solar energy

If there is a lack of demand, it's purely a result of willful manipulation
of the market by those whose interests are hurt by solar energy.

Yuri [45]3 years ago

4 0

The answer is the first three.

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3 years ago

The displacement of a wave traveling in the positive x-direction is:

rodikova [14]

Answer:

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Explanation:

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3 years ago

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Softa [21]

Chemical energy.

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3 years ago

Read 2 more answers

The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

Mademuasel [1]

Answer:

$28.11m$ far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

$(1). \: \:\beta =10 log(\dfrac{I}{I_0})$

The ratio of the intensities can be written as

$\frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} }$

$\dfrac{I}{I_0} = \dfrac{A_0}{A}.$

And since

$A = 4\pi r^2$

and

$A_0 = 4 \pi r_0^2$ ,

$\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2} = \dfrac{r_0^2}{r^2}$

meaning

$\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.$

Putting this into equation (1), we get:

$\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}$

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

$100dB=10 log(\dfrac{r_02}{(5m)^2})$

$10= log(\dfrac{r_0^2}{25})$

by taking both sides to the exponent:

$10^{10}= \dfrac{r_0^2}{25}$

$r^2 = 25 *10^{10}\\r = 5 *10^5$

Now equation (2) becomes

$\beta = 10log(\dfrac{25*10^{10}}{r^2})$

when the intensity level is 85 dB we have

$85 = 10log(\dfrac{25*10^{10}}{r^2})$

$8.5 = log(\dfrac{25*10^{10}}{r^2})$

take both sides to exponents and we get:

$10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}$

$10^{8.5} =\dfrac{25*10^{10}}{r^2}$

$r^2 = \dfrac{25*10^{10}}{10^{8.5}}$

$\boxed{r = 28.11m}$

Thus, $28.11m$ far from the speaker the intensity drops to 85 dB.

7 0

3 years ago

A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0

4 years ago