Which chemical equation represents a decomposition reaction? HELP ASAP!!!!

A one-dimensional plane wall of thickness 2l= 100 mm experiences uniform thermal energy generation of q˙= 800 w/m3 and is convec

slega [8]

Answer:

The thermal conductivity of the wall = 40W/m.C

h = 10 W/m^2.C

Explanation:

The heat conduction equation is given by:

d^2T/ dx^2 + egen/ K = 0

The thermal conductivity of the wall can be calculated using:

K = egen/ 2a = 800/2×10

K = 800/20 = 40W/m.C

Applying energy balance at the wall surface

"qL = "qconv

-K = (dT/dx)L = h (TL - Tinfinity)

The convention heat transfer coefficient will be:

h = -k × (-2aL)/ (TL - Tinfinty)

h = ( 2× 40 × 10 × 0.05) / (30-26)

h = 40/4 = 10W/m^2.C

From the given temperature distribution

t(x) = 10 (L^2-X^2) + 30 = 30°

T(L) = ( L^2- L^2) + 30 = 30°

dT/ dx = -2aL

d^2T/ dx^2 = - 2a

4 0

3 years ago

2. How do the phytochemicals present in various foods help us?<br>

vredina [299]

<h2>Answer:</h2>

Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.

8 0

3 years ago

A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d

QveST [7]

Answer:

The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

(target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

(10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

(target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

(target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

-94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

(12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

√(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0

3 years ago

A child and sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed o

Furkat [3]

<span> Using conservation of energy

Potential Energy (Before) = Kinetic Energy (After)

mgh = 0.5mv^2

divide both sides by m

gh = 0.5v^2

h = (0.5V^2)/g

h = (0.5*2.2^2)/9.81

h = 0.25m

</span>

4 0

3 years ago

Read 2 more answers

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.