What types of landforms can water form?

An energy storage system based on a flywheel (a rotating disk) can store a maximum of 3.7 MJ when the flywheel is rotating at 16

Likurg_2 [28]

The moment of inertia of the flywheel is 2.63 kg- $m^{2}$

It is given that,

The maximum energy stored on the flywheel is given as

E=3.7MJ= 3.7× $10^{6}$ J

Angular velocity of the flywheel is 16000 $\frac{rev}{min}$ = 1675.51 $\frac{rad}{sec}$

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

E = $\frac{1}{2}$ $Iw^{2}$

By rearranging the equation:

I = $\frac{2E}{w_{2} }$

I = 2.63 kg- $m^{2}$

Thus the moment of inertia of the flywheel is 2.63 kg- $m^{2}$ .

Learn more about moment of inertia here;

brainly.com/question/13449336

#SPJ4

7 0

1 year ago

A swimmer swims at 5 m/s. How long would it take to swim 5 laps of a 50m pool?

Gre4nikov [31]

About 5 hours gooood luck

3 0

1 year ago

How much potential energy does a 50-N box have when lifted at a height of 1.5M?

nikitadnepr [17]

The correct answer is: Option (A) 75 J

Explanation:

First, be careful with the units here. As you can see it is mentioned that there is a 50N box. It means that the weight (mg) of the box is given as the unit is Newton, not its mass (which is in kg).

As,

Potential-energy = mass * acceleration-due-to-gravity * height

PE = m*g*h --- (A)

In equation (A), mg is actually the weight of the box, which is given.

mg = 50N

h = height = 1.5m

Plug the values in equation (A):

PE = 50 * 1.5 = 75 J (Option A)

3 0

3 years ago

A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat

rewona [7]

Answer:

$\dot{m_{2}}=0.865 kg/s$

Explanation:

$\dot{m_1}= 0.5kg/s$

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

$\dot{m_{in}}=\dot{m_{out}}$

$\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}$

according to energy balance equation

$\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s$

4 0

3 years ago

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .