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3 years ago

What is the distance covered by a Freely falling object 5 seconds after being dropped ? After 6 seconds?

1 answer:

5 0

This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:

D = 1/2 A T²

Distance = (1/2) (acceleration) (time²)

The reason I never forgot it is because it's SO useful SO often. You really should memorize it. And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)

The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth. Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .

In 5 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (5 sec)²

D = (4.9 m/s²) (25 sec²)

D = 122.5 meters

In 6 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (6 sec)²

D = (4.9 m/s²) (36 sec²)

D = 176 meters

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calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s

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2 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

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3 years ago

Compare and contrast the life cycle of a low mass star and a high mass star

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Very High Mass: Run through their fuel exceedingly fast. *Die* relatively quickly (in the range of tens to hundreds of millions of years instead of billions and beyond) and go out with style, Supernova that will leave behind a neutron star (the *kind of very high mass stars" end this way) or a black hole (the *very very high mass stars* end this way.)

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3 years ago