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3 years ago

If there's air in a ball and it bounces how come a room doesn't bounce

Physics

1 answer:

finlep [7]3 years ago

4 0

Answer:

Funny answer but works!

Explanation:

Not enough air. Go outside, crack open the window and blow air into the room like you would a balloon then close the window quickly. If it doesn't start bouncing, try giving the house a push. Good luck.

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A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr

Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0

3 years ago

A bar magnet is held in place while another bar magnet is placed near it. The second bar magnet spins around and

Vinil7 [7]

Answer:

Explanation:

the increase energy stored in thw system is proportional to the decrease in kinetic energy

4 0

3 years ago

Pablo's engineering team has developed a new material that is strong enough to withstand major impacts, including bullets, witho

marin [14]

Answer:

the answer is c

Explanation:

4 0

3 years ago

NEED HELP ASAP!!!!

podryga [215]

The answer is B
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6 0

3 years ago

a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/

LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

$0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\$

we isolate th (needed to reach the maximum height hmax)

$th = \frac{v_{0}*sin(\alpha)}{g}$

The formula describing vertical distance is:

$y = Vy * t-g* t^{2} / 2$

So, given y = hmax and t = th, we can join those two equations together:

$hmax = Vy* th-g*th^{2}/2$

$hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)$

if we launch a projectile from some initial height h all you need to do is add this initial elevation

$hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)$

$hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft$

6 0

3 years ago