an airplane flies 1,592 miles east from arizona to georgia in 3.68 hours. what is the average velocity of the airplane

Riders in a carnival ride stand with their backs against the wall of a circular room of diameter

Veseljchak [2.6K]

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

= $45 \times \dfrac{2\pi}{60}$

=4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

= 88.83m x μ

now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

μ = 0.11

hence, the correct answer is option C

6 0

4 years ago

A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axle. A second cylinder, with m

MAXImum [283]

Answer:

Part(a): The final angular velocity is $\omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}$

Part(b): The ratio of the rotational energies is $\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}$ ,showing the the energy of th system will decrease.

Explanation:

Part(a):

If ' $I$ ' be the moment of inertia of an object and ' $\omega$ ' be its angular velocity then the angular momentum ' $L$ ' of the object can be written as

$L = I \omega$

If ' $I_{1}$ ' and ' $I_{2}$ ' be the moment of inertia of the two cylinders and ' $\omega_{1}$ ' and ' $\omega_{2}$ ' be the initial angular velocity of the cylinders and ' $\omega_{1}{'}$ ' and ' $\omega_{2}^'}$ ' be their respective final angular velocity, then from conservation of angular momentum,

$I_{1} \omega_{1} + I_{2} \omega_{2} = I_{1} \omega_{1}^{'} + I_{2} \omega_{2}^{'}$

Given, $\omega_{1} = \omega_{i},~\omega_{2} = 0,~\omega_{1}^{'} = \omega_{2}^{'} = \omega_{f}$ . From the above expression

$&& I_{1} \omega_{i} = (I_{1} + I_{2}) \omega_{f}\\&or,& \omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}$

Part(b):

Initial kinetic energy

$K_{i} = \dfrac{1}{2} I_{1} \omega_{i}^{2}$

and Final kinetic energy

$K_{f} = \dfrac{1}{2}(I_{1} + I_{2}) \omega_{f}^{2}$

Substituting the value of $\omega_{f}$ ,

$&& K_{f} = \dfrac{1}{2}(I_{1} + I_{2})\dfrac{I_{1}^{2}\omega_{i}^{2}}{(I_{1} + I_{2})^{2}} = \dfrac{1}{(I_{1} + I_{2})} \dfrac{1}{2}I_{1}\omega_{i}^{2} = \dfrac{1}{(I_{1} + I_{2})} K_{i}\\&\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}$

The above expression shows that the ebergy of the system will decrease.

7 0

3 years ago

According to this graph, which statement describes how the amount of iron changes over time in the reaction

VashaNatasha [74]

The graph shows the production of Fe, from the graph that it increases rapidly and then slowly increases.<span>The answer is a! (:
</span>

5 0

4 years ago

How high up is a 3 kg object that has 300 joules of energy

kvasek [131]

Answer:

Height, h = 10.20 meters

Explanation:

It is given that,

Mass of the object, m = 3 kg

Energy of object, E = 300 J

Let it will moved to a height of h. The energy possessed by it is called gravitational potential energy. It is given by :

$E=mgh$

$h=\dfrac{E}{mg}$

$h=\dfrac{300\ J}{3\ kg\times 9.8\ m/s^2}$

h = 10.20 meters

So, the object will move to height of 10.20 meters. Hence, this is the required solution.

5 0

4 years ago

Read 2 more answers

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitationa

sdas [7]

Answer:

The correct answer is option B)

Explanation:

Considering the given question as -

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitational pull on the shuttle? A) The moon pulls more on the shuttle. B) The earth pulls more on the shuttle. C) Both are equal due to equal distances. D) Both are equal due to the mass of the shuttle.

We know that gravitational pull (F) between any two bodies of mass $M_{1}$ and $M_{2}$ is given by -