Ask question

Ask question

All categories

3 years ago

11

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 ???? 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:

sladkih [1.3K]3 years ago

5 0

Answer:

25.8 rad/s

Explanation:

C = circumference of the hollow sphere = 0.749 m

r = radius of the sphere

Circumference of the hollow sphere is given as

C = 2π r

0.749 = 2 (3.14) r

r = 0.12 m

m = mass of the basketball = 0.624 kg

Moment of inertia of basketball is given as

$I = \left ( \frac{2}{3} \right )mr^{2}$

$I = \left ( \frac{2}{3} \right )(0.624)(0.12)^{2}$

I = 5.99 x 10⁻³ kg-m²

w = angular speed

KE = Kinetic energy of the ball = 1.99 J

Kinetic energy of the ball is given as

KE = (0.5) I w²

1.99 = (0.5) (5.99 x 10⁻³) w²

w = 25.8 rad/s

You might be interested in

Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at

klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

$\lambda$ = Wavelength

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

Energy of a photon is given by

$E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}$

Let $\lambda_1$ = 700 nm

$\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}$

For red light

$E_1=\dfrac{hc}{\lambda_1}$

For UV light

$E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}$

Dividing the equations

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1$

Hence, the answer is b) twice the energy of each photon of the red light.

7 0

3 years ago

Read 2 more answers

What is the momentum of a man of mass 10kg when he walks with a uniform velocity of 2m/s.

Stolb23 [73]

Answer:

$Momentum(p) = 20kgm/s$

Explanation:

Given

Mass = 10kg

Velocity = 2m/s

Required

Calculate the momentum of the man

Momentum is calculated as thus

$Momentum(p) = Mass(m) * Velocity(v)$

or

$p = mv$

So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;

The formula becomes

$Momentum(p) = 10kg * 2m/s$

$Momentum(p) = 10 * 2 * kg * m/s$

$Momentum(p) = 20kgm/s$

Hence, the momentum of the man is $20kgm/s$

5 0

3 years ago

Can someone help me wit hthis science question?

klemol [59]

Answer:

see explanation

Explanation:

There is an increasing demand for materials and natural resources from a growing global population, especially those in more economically developed countries. The world's resources are being used up more quickly. The consumption of resources is spread unequally between MEDCs (more economically developed countries), who use more resources, and LEDCs (less economically developed countries), who use less.

The gap between the rich and poor is more evident when the resources are shared so unevenly and unfairly and natural resources like materials and natural energy cannot reach the demand of the people which can have consequences and be very difficult to manage. Having a lack of these materials in a country can result in prices going up for them, and the industry could be harder to work in because of a lack of materials.

6 0

3 years ago

a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em> has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

$W=Fx$

This is equal to the change in the bullet's kinetic energy.

$W=Fx=\frac{1}{2} m(u^2-v^2)......(1)$

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

$F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)$

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

$\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s$

Thus the speed of the bullet is 71 m/s

3 0

3 years ago

How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have

cricket20 [7]

1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
$E=mgh=(80 kg)(9.81 m/s^2)(5 m)=3924 J$
So the power output is
$P= \frac{E}{t}= \frac{3924 J}{30 s} \sim 130 W$

And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.

2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.

6 0

3 years ago

Read 2 more answers