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shtirl [24]
1 year ago
13

If you drop an object, it will accelerate downward at a rate of 9.8 meters per second per second. if you instead throw it downwa

rds, its acceleration (in the absence of air resistance) will be?
Physics
1 answer:
Alla [95]1 year ago
6 0

In the absence of air resistance or any other force on it, an object's acceleration is the acceleration of gravity ... <em>9.8 m/s² downward</em>.

It doesn't matter if the object rolled off a table, or fell out of an eagle's mouth, or got shot straight up out of a cannon, or LeBron threw it at the basket for a 3-pointer, or Justin Tucker kicked it for a field goal, or Tiger Woods chipped it for a bogey, or Billy Joe MacAllister heaved it straight down off the Tallahatchie Bridge with all his strength.

It doesn't matter how it got STARTED moving, or how fast it's going right now, or what direction it's going now.

If there's no air resistance, and no OTHER force on the object that can speed it up or slow it down, then it's in "free fall", and it's accelerating at 9.8 m/s² downward.

The size and direction of its MOTION doesn't matter. But the size and direction of its acceleration is 9.8 m/s² straight down.  

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1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame
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Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0
3 years ago
IN WHAT CONDITION DO SOUND ECHO
DerKrebs [107]

Answer:

The conditions necessary for hearing the echo. The distance between the sound source and the reflecting surface must not be less than 17 metres where the time period between hearing the original sound and its echo should not be less than 0.1 of a second.

6 0
2 years ago
An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in bet
Vera_Pavlovna [14]

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

4 0
3 years ago
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