Ask question

Ask question

All categories

3 years ago

14

A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 x 10–3 kg∙m2 is suspended from the ceiling. A rope passes over

it with a 2.0-kg block attached to one end and a 4.0-kg block attached to the other. The rope does not slip on the pulley. When the velocity of the heavier block is 2.0 m/s the total kinetic energy of the pulley and blocks is:

1 answer:

vekshin13 years ago

6 0

Answer:

22J

Explanation:

Given :

radius 'r'= 3cm

rotational inertia 'I'=4.5 x $10^{-3}$ kgm²

mass on one side of rope ' $m_{1$ '= 2kg

mass on other side of rope' $m_{2$ ' =4kg

velocity'v' of mass $m_{2$ ' = 2m/s

Angular velocity of the pulley is given by

‎ω = v /r => 2/ 3x $10^{-2$

‎ω = 66.67 rad/s

For the rotating body, we have

KE = $\frac{1}{2}$ I ω²

$KE_p = \frac{1}{2} (4.5 *10^{-3} )(66.67^{2} )$

$KE_p$ = 10J

Next is to calculate kinetic energy of the blocks :

$KE_{b} = \frac{1}{2} (m_1 + m_2).v^2\\KE_b= \frac{1}{2} (2+4).2^2$

$KE_b$ =12J

Therefore, the total kinetic energy will be

KE = $KE_p + KE_b$ =10 + 12

KE= 22J

You might be interested in

Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of t

Viefleur [7K]

It’s is D bc it is and idk you should do it

6 0

3 years ago

When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g

Olegator [25]

Answer:

Your question was incomplete so here is the complete question and answer.

Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)

a) plain water

b) 5-7 percent glucose solution

c) Glucose polymer solution of 6-8 percent

d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Explanation:

Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.

3 0

3 years ago

HELP ME PLEASEEEEEEEEEEEEEE

ozzi

Answer: The correct statements are:

The atoms are very attracted to one another.
The atoms are held tightly together.

Explanation:

Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.

In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.

The key point about solid are:

The atoms are very attracted to one another.
The atoms are not moving freely.
It will not spread out evenly to fill any container.
The atoms are held tightly together.
The forces of attraction are strong to bring molecules together.
The atoms are close and in fixed positions.

7 0

3 years ago

1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown

denis23 [38]

Answer:

a) 3.37 x $10^{3} kg/m^3$

b) 6.42kg/ $m^{3}$

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x $10^{-4}$ $kg/m^{3}$ . So density of metal = mass of metal / volume of metal = 5 / 14 x $10^{-4}$ $kg/m^{3}$ = 3.37 x $10^{3} kg/m^3$

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/ $m^{3}$

3 0

2 years ago

Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above

poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$W_{dis}$ - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

$f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}$

If $m = 1000\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 40\,m$ , $y_{2} = 25\,m$ and $\Delta s = 400\,m$ , then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0

3 years ago