All categories

3 years ago

In a game of pool, why will the balls eventually stop after a collision?

1 answer:

7 0

As a ball rolls ... even if there were no air resistance ... the ball
has to push down and climb over every one of those little fibers
that stick up from the felt fabric on the table. As it does that, each
little fiber brushes against the ball, and robs a tiny tiny bit of its
kinetic energy.

You might be interested in

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box= $u=13m/s$

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

$v^2-u^2=2as$

Substitute the values

$(11.5)^2-(13)^2=2a(8.5)$

$132.25-169=17a$

$-36.75=17a$

$a=\frac{-36.75}{17}=-2.2m/s^2$

We know that

Acceleration= $a=\frac{v-u}{t}$

Substitute the values

$-2.2=\frac{11.5-13}{t}$

$-2.2=\frac{-1.5}{t}$

$t=\frac{1.5}{2.2}=0.68 s$

Hence, the time taken by box to slow by this amount=0.68 s

8 0

3 years ago

A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, wi

Vitek1552 [10]

Answer:

Explanation:

Let the velocity be v

Total energy at the bottom

= rotational + linear kinetic energy

= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell = mr² )

= 1/2 mr²ω² + 1/2 mv² ( v = ω r )

= 1/2 mv² +1/2 mv²

= mv²

mv² = mgh ( conservation of energy )

v² = gh

v = √gh

= √9.8 x 1.8

= 4.2 m /s

8 0

3 years ago

A convex lens of focal length 35 cm produces a magnified image 2.5 times the size of the object. What is the object distance if

Zanzabum

Answer:

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Explanation:

u = Object distance

v = Image distance

f = Focal length = 35

m = Magnification = 2.5

$m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u$

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm$

$v=-2.5\times 21=-52.5\ cm$

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

3 0

3 years ago

A man walks 2.6 miles east, then turns and walks 6.7 miles north. What is his resultant

Zarrin [17]

Answer:

Yes cause he walks 6.7 miles

6 0

3 years ago

The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top

vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

$v_y^2 =v_0^2 +2ay$

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

$v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}$

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s $v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m$

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

7 0

4 years ago