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3 years ago

15

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.99 J of work to set the ball spinnin

g from rest, at what angular speed ω will the ball rotate? Model a basketball as a thin-walled hollow sphere. For a men's basketball, the ball has a circumference of 0.749 m and a mass of 0.624 kg .

1 answer:

scoundrel [369]3 years ago

8 0

To solve this problem it is necessary to apply the concepts related to rotational kinetic energy, the definition of the moment of inertia for a sphere and the obtaining of the radius through the circumference. Mathematically kinetic energy can be given as:

$KE= I\omega^2$

Where,

I = Moment of inertia

$\omega =$ Angular velocity

According to the information given we have that the radius is

$\Phi= 2\pi r$

$0.749m = 2\pi r$

$r = 0.1192m$

With the radius obtained we can calculate the moment of inertia which is

$I = \frac{2}{3}mr^2$

$I = \frac{2}{3}(0.624)(0.1192)^2$

$I = 5.91*10^{-3} kg \cdot m^2$

Finally, from the energy equation and rearranging the expression to obtain the angular velocity we have to

$\omega = \sqrt{\frac{2KE}{I}}$

$\omega = \sqrt{\frac{2(1.99)}{5.91*10^{-3}}}$

$\omega = 25.95rad/s$

Therefore the angular speed will the ball rotate is 25.95rad/s

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Effciency of a lever is never 100% or more. why?Give reason

Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below $100\%$ .

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let $D_\text{in}$ and $D_\text{out}$ denote these two distances, and let $F_\text{in}$ and $F_\text{out}$ denote the input and the output forces. If the lever is indeed idea, then:

$F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}$ .

Rearrange to obtain:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}$

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means $D_\text{in} > D_\text{out}$ , such that $F_\text{in} < F_\text{out}$ .

Despite $F_\text{in} < F_\text{out}$ , the amount of work required will stay the same. Let $s_\text{out}$ denote the required linear displacement for the output force. At a distance of $D_\text{out}$ from the fulcrum, the angular displacement of the output force would be $\displaystyle \frac{s_\text{out}}{D_\text{out}}$ . Let $s_\text{in}$ denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be $\displaystyle \frac{s_\text{in}}{D_\text{in}}$ . Because both the input and the output are on the same lever, their angular displacement should be the same:

$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

Rearrange to obtain:

$\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}$ .

While increasing $D_\text{in}$ reduce the size of the input force $F_\text{in}$ , doing so would also increase the linear distance of the input force $s_\text{in}$ . In other words, $F_\text{in}$ will have to move across a longer linear distance in order to move $F_\text{out}$ by the same $s_\text{out}$ .

The amount of work required depends on both the size of the force and the distance traveled. Let $W_\text{in}$ and $W_\text{out}$ denote the input and output work. For this ideal lever:

$\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}$ .

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be $100\%$ - not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, $F_\text{in}$ would be larger than ideal:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance})$ .

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below $100\%$ ,

4 0

3 years ago

I need help on (a)<br> I don't know what equation to use?

Alchen [17]

Impulse = (force) x (length of time the force lasts)

I see where you doodled (60)(40) over on the side, and you'll be delighted
to know that you're on the right track !

Here's the mind-blower, which I'll bet you never thought of:
On a force-time graph, impulse (also change in momentum)
is just the <em>area that's added under the graph during some time</em> !

From zero to 60, the impulse is just the area of that right triangle
under the graph. The base of the triangle is 60 seconds. The
height of the triangle is 40N . The area of the triangle is not
the whole (base x height), but only <em><u>1/2 </u></em>(base x height).

1/2 (base x height) = 1/2 (60s x 40N) = <u>1,200 newton-seconds</u>

<u>That's</u> the impulse during the first 60 seconds. It's also the change in
the car's momentum during the first 60 seconds.

Momentum = (mass) x (speed)

If the car wasn't moving at all when the graph began, then its momentum is 1,200 newton-sec after 60 seconds. Through the convenience of the SI system of units, 1,200 newton-sec is exactly the same thing as 1,200 kg-m/s . The car's mass is 3 kg, so after 60 sec, you can write

Momentum = M x V = (3 kg) x (speed) = 1,200 kg-m/s

and the car's speed falls right out of that.

From 60to 120 sec, the change in momentum is the added area of that
extra right triangle on top ... it's 60sec wide and only 20N high. Calculate
its area, that's the additional impulse in the 2nd minute, which is also the
increase in momentum, and that'll give you the change in speed.

8 0

3 years ago

HELP URGENT!!!!!!!!!!!!!!!!!!!!!!!!

nirvana33 [79]

D
Because the rest of the answers are illogical

5 0

3 years ago

A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi

svetlana [45]

Answer:

$E=0$ at r < R;

$E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}$ at 2R > r > R;

$E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}$ at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

$\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}$ (integral over a closed surface)

where,

$E$ = Electric field

$Q_{enclosed}$ = charged enclosed within the closed surface

$\epsilon$ = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

$Q_{enclosed}$ = 0 and hence $E$ = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

$Q_{enclosed}$ = Q,

therefore,

$E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}$

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to $4\pi r^{2}$ )

or, $E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}$

at r >= 2R

$Q_{enclosed}$ = 2Q

Hence, by similar calculations, we get,

$E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}$

4 0

4 years ago

A ball starts at rest and rolls down a ramp, developing a speed of 45 m/s in 7 seconds. Calculate the ball's acceleration.

Alexxx [7]

Answer:

6.429 m/s^2.

Explanation:

Using equations of motion,

i. vf = vi + at

ii. vf^2 = vi^2 + 2a*S

iii. S = vi*t + 1/2 * (a*t^2)

Where,

vf = final velocity of the motion

vi = initial velocity of the motion

S = distance travelled

t = time taken to complete the motion

a = acceleration due to gravity

Given:

vi = 0m/s

vf = 45 m/s

t = 7 s

a = ?

Using the i. equation of motion,

vf = vi + at

45 = 0 + a*7

a = 45/7

= 6.429 m/s^2

6 0

3 years ago