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Irina-Kira [14]
2 years ago
9

If someone were to drop an orb from a height of 34,000 meters approximately, and then drop a second orb at 34,100 meters, estima

te how long it would take for the second orb to reach the ground if it takes the first orb 55~ seconds to impact.
Extra Info:
-The first orb weights 2 pounds and the second orb weights only 2.5 pounds.
-The ground is entirely flat and level. (as if you smashed a flattening board into the ground)
-They are both dropped with no force put into it, simply held up and let go of.
-Both Orbs are the same size, just different density.
-The heights are estimates and are not exact, hence the answer can be an estimated one.

(this question sparked from boredom ._.)
If info required I can provide.
Physics
1 answer:
Pavlova-9 [17]2 years ago
5 0

Answer:

Explanation:

SO the ball falls 618 meters per second meaning that's how fast 2 pounds fall per second so i have to find out how fast .5 pounds falls which i think its 154 meters per second per .5 pounds meaning the second orb would fall 772 meters per second so the second orb would take 44.1 seconds to hit on impact. Im not sure if this is right but this is my best guess.

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Answer:

hi there!

the correct answer to this question is: 6.67 mph

Explanation:

you convert minutes to hours

10 miles * 60 mins / 90 mins

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3 years ago
What is the use of an inclined plane​
loris [4]

Answer:

Hey there!

Inclined planes are used to lift heavy objects to higher places.

Hope this helps :)

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3 years ago
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One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau
suter [353]

Answer:

actual efficiency is  47.78 %

Carnot efficiency  is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency  and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy  = 7.0 × 10^9 cal  (4.184 J/1 cal)  = 29.29 × 10^9 J

actual efficiency  =  1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency  =  0.4778

actual efficiency is  47.78 %

and

Carnot efficiency  is = 1 - ( 703 / 2173 )

so Carnot efficiency  is  = 0.67648

Carnot efficiency  is 67.65 %

and

power output  = work / time

power output  =  1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0
3 years ago
In which direction does the sun appear to move across the sky
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6 0
3 years ago
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What quantity of heat must be removed from 20g<br>of water at 0°C to change it to ice at 0°C?​
seraphim [82]

The quantity of heat must be removed is 1600 cal or 1,6 kcal.

<h3>Explanation : </h3>

From the question we will know if the condition of ice is at the latent point. So, the heat level not affect the temperature, but it can change the object existence. So, for the formula we can use.

\boxed {\bold {Q = m \times L}}

If :

  • Q = heat of latent (cal or J )
  • m = mass of the thing (g or kg)
  • L = latent coefficient (cal/g or J/kg)
<h3>Steps : </h3>

If :

  • m = mass of water = 20 g => its easier if we use kal/g°C
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Q = ... ?

Answer :

Q = m \times L \\ Q = 20 \times 80 = 1600 \: cal

So, the quantity of heat must be removed is 1600 cal or 1,6 kcal.

<u>Subject : Physics </u>

<u>Subject : Physics Keyword : Heat of latent</u>

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