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shtirl [24]
2 years 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]2 years 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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3 years ago
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4 years ago
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length t
gladu [14]

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

$t=\sqrt{\frac{2H}{g}}$

v_1t= 0.35  ...........(i)

v_0t= 1.09  ...........(ii)

Dividing the equation (ii) by (i)

$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$

∴ v_0=3.11  \ v_1

Now loss of energy  = change in the kinetic energy

$W=\frac{1}{2} m [v_0^2-v_1^2]$

$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$

$W=7.307\times 10^{-3} \ v_0^2$

If f is average friction force, then

(f)(L) = W

(f) (1) = $7.307\times 10^{-3} \ v_0^2$

(f)  = $7.307\times 10^{-3} \ v_0^2$

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3 years ago
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A charged particle has an electrostatic field surrounding it.

When the particle is in motion, the moving charge is an
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