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3 years ago

13

How does a car batteray NOT die?

1 answer:

erik [133]3 years ago

6 0

Answer:

bye hooking plugs up to it to amp it up

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Though still is on the margins of the global agro-food system, organic farming practices are on the rise in the core. In growing

lisabon 2012 [21]

Answer: composted cow manure

Explanation:

A cow manure consists of high levels of ammonia, pathogens which can harm the plants thus cannot be used as fertilizer. The composted cow manure is a processed fertilizer which is prepared by eliminating the ammonia gas and pathogens. It is prepared by adding up the additional organic matter.

It is an excellent growing medium for the agricultural crops and garden plants as it is a nutrient rich fertilizer. It is mixed into the soil or can be used for top dressing of the agricultural field. It is a kind of organic manure.

4 0

2 years ago

Por favor. alguien me comunique con fatimalisethmateual

lions [1.4K]

Wait why do you want me to

7 0

2 years ago

A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I h

anygoal [31]

Answer: Attached below is the missing diagram

answer :

A) 1) Wr > WI, 2) Qc' > Qc

B) 1) QH' > QH, 2) Qc' > Qc

Explanation:

л = w / QH = 1 - Qc / QH and QH = w + Qc

<u>A) each cycle receives same amount of energy by heat transfer</u>

<u>(</u> Given that ; Л1 = 1/3 ЛR )

<em>1) develops greater bet work </em>

WR develops greater work ( i.e. Wr > WI )

<em>2) discharges greater energy by heat transfer</em>

Qc' > Qc

solution attached below

<u>B) If Each cycle develops the same net work </u>

<em>1) Receives greater net energy by heat transfer from hot reservoir</em>

QH' > QH ( solution is attached below )

<em>2) discharges greater energy by heat transfer to the cold reservoir</em>

Qc' > Qc

solution attached below

4 0

2 years ago

How is a scale model different from other types of models?

SVETLANKA909090 [29]

Answer:

a scale model each size is a certain amount smaller

Explanation:

6 0

3 years ago

The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm

Angelina_Jolie [31]

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= $\frac{38}{0.25}$

= $152 \ layers$

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= $38\times 38 \ mm^2$

= $1444 \ mm^2$

The area covered every \second will be:

= $d\times v$

= $1.25\times 40$

= $50 \ mm^2$

The time required to deposit one layer will be:

= $\frac{1444}{50}$

= $28.88 \ sec$

The time required for one layer will be:

= $15 \ sec$

∴ Total times required for one layer will be:

= $15+28.88$

= $43.88 \ sec$

So,

Number of layers = 152

Therefore,

Total time will be:

= $152\times 43.88$

= $6669.76 \ sec$

= $1 \ hour \ 51 \ min$

6 0

3 years ago