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Evgesh-ka [11]
4 years ago
8

Why can’t the high-quality energy released from a lump of burning coal be recycled?

Physics
1 answer:
snow_tiger [21]4 years ago
5 0

Answer:

when energy changes from one form to another, it always goes from a more useful to a less useful form. This means that people can't recycle or reuse high quality energy to perform useful work.

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You watch distant Sally Homemaker driving nails into a front porch at a regular rate of 1 stroke per second. You hear the sound
san4es73 [151]

Answer:

d = 343 m

Explanation:

As we know that the speed of sound in air is given as

v = 343 m/s

now the rate of strike is 1 stroke per second

so distance is given as

d = v t

now we will have

t = 1 s

d = 343 \times 1

so the distance is given as

d = 343 m

5 0
3 years ago
What is the formula for silver nitrate
serious [3.7K]

Answer:

Explanation:

Agno3

8 0
4 years ago
A crate of 890 kg gets raised to a height of 2.1m. Calculate the crate's gravitational potential energy.​
mars1129 [50]

Answer:

18,316.2

Explanation:

The formula for GPE is mgh, where

M = Mass of the object

G = Acceleration due to gravity (9.8 m/s^2 on Earth)

H = Height above ground

890 × 9.8 × 2.1 = 18,316.2

4 0
3 years ago
Read 2 more answers
A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.
noname [10]

Answer:

The value is  \rho_s  =  4.026 *10^{-6} \  C/m^2

Explanation:

From the question we are told that

   The radius of the inner conductor  is  r_1 = 1 \ mm =  0.001 \ m

    The radius of the outer conductor is  r_2 = 3 \ mm = 0.003 \  m

    The potential at the outer conductor is  V = 1.5 kV  =  1.5 *10^{3} \  V

Generally the capacitance per length of the capacitor like set up of the two conductors is

      C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}

Here \epsilon_o is the permitivity of free space with value  \epsilon_o =  8.85*10^{-12} C/(V \cdot m)

=>   C= \frac{2 *  3.142  * 8.85*10^{-12}  }{ ln [\frac{0.003}{0.001} ]}

=>   C= 50.6 *10^{-12} \  F/m

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      \rho_l  =  C *  V

=>   \rho_d  =  50.6*10^{-12} *  1.5*10^{3}

=>   \rho_d  =  7.59*10^{-8} \  C/m

Generally the surface charge density is mathematically represented as

        \rho_s  =  \frac{\rho_l }{2 \pi * r_2 }    here 2 \pi r = (circumference \ of \ outer \  conductor  )

=>    \rho_s  =  \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }

=>    \rho_s  =  4.026 *10^{-6} \  C/m^2

3 0
3 years ago
Which of the following is equivalent to 10 kilograms?
EleoNora [17]

a) 1000 grams = 1 kg ... no


d) 1000 centigrams


7 0
3 years ago
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