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Serhud [2]
3 years ago
10

someone throws a rubber ball vertically upward from the roof of a building 8.2m in height. the ball rises, then falls.it just mi

sses the edge of the roof, and strikes the ground.if the ball is in the air for 6.2s, what was its initial velocity?(disregard air resistance. a=-g=-10m/s²
Physics
1 answer:
laila [671]3 years ago
5 0

Answer:

H = V t - 1/2 g t^2         V and g are in different directions

-8.2 = V * 6.2 - 1/2 * 10 * 6.2*2 = 6.2 V - 192.2

V = 184 / 6.2 = 29.7 m/s

Check: find time for ball to return to zero height

0 = 29.7 T - 5 T^2

T = 29.7 / 5 = 5.94 sec

The ball must have fallen 8.2 m in (6.2 - 5.94) sec = .26 sec

S = 29.7 * .26 + 5 * .26^2 = 8.1

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A truck accelerates from a stop to 27m/sec in 9 minutes. What was the trucks acceleration?
olchik [2.2K]
The acceleration is 3 m/s per minute, or 0.05 m/s per second.
5 0
4 years ago
Which of the following is an example of a force?
strojnjashka [21]

The answer would be d!

6 0
3 years ago
A 49.3 g ball of copper has a net charge of 2.0 µc. what fraction of the copper's electrons has been removed? (each copper atom
Serggg [28]
First, find how many copper atoms make up the ball: 
 moles of atoms = (49.3 g) / (63.5 g per mol of atoms) = 0.<span>77638</span><span>mol 
</span> # of atoms = (0.77638 mol) (6.02 × 10^23 atoms per mol) = 4.6738*10^23<span> atoms </span>

<span> There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom:
</span> normal # electrons = (4.6738 × 10^23 atoms) (29 electrons per atom) = <span> <span>1.3554</span></span><span>× 10^25 electrons 
</span>
<span> Currently, the charge in the ball is 2.0 µC, which means -2.0 µC worth of electrons have been removed.
</span><span> # removed electrons = (-2.0 µC) / (1.602 × 10^-13 µC per electron) = 1.2484 × 10^13 electrons removed
 
</span><span> # removed electrons / normal # electrons = </span>
<span>(1.2484 × 10^13 electrons removed) / (1.3554 × 10^25 electrons) = 9.21 × 10^-13 </span>

<span> That's 1 / 9.21 × 10^13 </span>
7 0
3 years ago
PART ONE
Lina20 [59]

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass.  Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

4 0
3 years ago
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