With increasing depth, the temperature within earth

The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its

STatiana [176]

Answer:

a. global warming

Explanation:

that's the definitain of global warming

6 0

2 years ago

What is the value of the thickness of the boundary layer at leading edge of the plate?

Harman [31]

Along the flow direction, the boundary layer's thickness varies. For ReT values between 260 and 780, the boundary layer thickness—defined as the depth at which the normalized concentration has a value of 1/e—ranges between 800 and 250 m.

<h3>What is the Boundary layer?</h3>

A boundary layer is the thin layer of fluid that forms immediately around a bounded surface in physics and fluid mechanics as a result of the fluid flowing along the surface.
A no-slip boundary condition is created as a result of the fluid and wall interaction (zero velocity at the wall).
After that, the flow velocity above the surface steadily rises until it reaches the bulk flow velocity again.
The term "velocity boundary layer" refers to the thin layer of fluid whose velocity has not yet recovered to that of the main flow.

<h3>What purpose does a boundary layer serve?</h3>

Because it is on the fluid's boundary, engineers refer to this layer as the boundary layer.
Many aerodynamics issues, such as wing stall, skin friction drag on an object, and the heat transfer that takes place in high-speed flight, depend heavily on the specifics of the flow within the boundary layer.

Learn more about Boundary layer here:

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7 0

1 year ago

In which scenario will the two objects have the greatest gravitational force

Brilliant_brown [7]

Answer: I think the answer C

Explanation:

7 0

2 years ago

Read 2 more answers

You carry a 5 kg sack of potatoes across the store for 5 minutes, and you wait in line holding it for 30 seconds before you get

ludmilkaskok [199]

Answer:

The only work done is when the person lifts the sack over a distance, W = 78.48 [N]

Explanation:

We have to remember the definition of work, which tells us that work is the result of a force by a distance, we must apply this concept in each of the movements of the person in the problem described.

W = F * d

where:

F = force [N]

d = distance [m]

The force is given by the producto of the mass by the gravity.

F = 5 * 9.81 = 49.05 [N]

W = 49.05 * 1.6 = 78.48 [N]

3 0

3 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = <em>6738.27 J</em>

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s