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Hatshy [7]
3 years ago
5

How long will it take an object to hit the ground if it is dropped from a hight of 176.4 meters ​

Physics
1 answer:
Oliga [24]3 years ago
5 0

There's a short handy formula for that.

If the object is just dropped and not tossed, and it's not affected by air resistance on the way down, then the distance it falls in T seconds is

D = (1/2) (gravity) (T²)

For this problem . . .

176.4 m = (1/2) (9.8 m/s²) (T²)

Divide each side by  (4.9 m/s²) :

T² = (176.4 m) / (4.9 m/s²)

T² = (36 s²)

Take the square root of each side:

<em>T = 6 seconds</em>

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Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

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Initial velocity V_{i} = 60 mph = 26.8224 m/s

Final velocity V_{f} = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = \frac{1}{2}m(  V_{i}² - V_{f}² )

we substitute

Δk = \frac{1}{2}×1814.37( (26.8224)² - (13.4112)² )

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Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

so

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0
3 years ago
The chart shows rate of decay. A 3 column table with 7 rows. The first column is Half-lives elapsed, with entries 0, 1, 2, 3, 4,
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By Kirchoff's law in left side loop

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similarly kirchoff's law in right side loop

E_3 - E_4 - E_2 = (r_3 + r_4 + R_3)I_2 - (R_2 + r_2)I_3

also by junction law we know that

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now by plug in all values we have

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