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3 years ago

Does the mass change how a bouncy ball bounce

1 answer:

6 0

Mass affects a ball's bounce through kinetic energy. The more mass an object has, the more kinetic energy it has when dropped, due to gravity. How much the ball deforms is based on its chemical makeup, or in this case, elasticity. When the ball deforms, the kinetic energy is converted into potential energy. As many kinds of balls have high amounts of elasticity, the potential energy converts back to kinetic energy when the deformation of the ball returns to its normal state. If the force of impact is too great for the ball to absorb, it may collapse and lose its bounce as the energy is dissipated in a different manner.

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Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago

What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k

oksano4ka [1.4K]

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N

4 0

3 years ago

Read 2 more answers

We are usually not aware of the electric force acting between two everyday objects because

tatuchka [14]

Answer:We are usually not aware of the electric force acting between two everyday objects because most everyday objects have as many plus charges as minus charges. Option A

Explanation:An electric force is exerted between any two charged objects( either positive or negative). Objects with the same charge will repel each other, and objects with opposite charge will attract each other. The strength of the electric force between any two charged objects depends on the amount of charge that each object contains and on the distance between the two charges. Electric charges are generated all around us due to different surfaces bearing different types of charges. We are usually not aware of it because the quantity of positive charges equals the number of negative charges.

3 0

3 years ago

Projectile motion in two dimensions cannot be determined by breaking the problem into two connected one-dimensional problems.

andriy [413]

The answer I believe is False

8 0

3 years ago

¿Un cuerpo que se encuentra a 5°C puede darle calor a otro cuerpo?

likoan [24]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

7 0

3 years ago