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3 years ago

Can someone explain C s +O2 -> CO g

Physics

1 answer:

Semmy [17]3 years ago

8 0

Explanation:

Correct equation is

C + O2 -> CO2

Carbon react with Oxygen to form carbon dioxide

This is a combustion reaction

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A jet airplane is in level flight. The mass of the airplane is m=9010kg. The airplane travels at a constant speed around a circu

Lyrx [107]

Answer:

The magnitude of the lift force L = 92.12 kN

The required angle is ≅ 16.35°

Explanation:

From the given information:

mass of the airplane = 9010 kg

radius of the airplane R = 9.77 mi

period T = 0.129 hours = (0.129 × 3600) secs

= 464.4 secs

The angular speed can be determined by using the expression:

ω = 2π / T

ω = 2 π/ 464.4

ω = 0.01353 rad/sec

The direction $\theta = tan^{-1} ( \dfrac{\omega ^2 R}{g})$

$\theta = tan^{-1} ( \dfrac{0.01353 ^2 \times (9.77\times 1609)}{9.81})$

θ = 16.35°

The magnitude of the lift force L = mg ÷ Cos(θ)

L = (9010 × 9.81) ÷ Cos(16.35)

L = 88388.1 ÷ 0.9596

L = 92109.32 N

L = 92.12 kN

3 0

3 years ago

Kyle is wondering what he is getting for his birthday. He decides to shake the box that contains his present. Kyle did not reali

stiks02 [169]

Answer: Asking questions

Explanation:

Scientists always start with their question before

Observing anything

8 0

3 years ago

Difine scalar quantity

vfiekz [6]

Scalar quantity are physical quantities that have just magnitude, not direction.

It is always positive.
Examples: Speed, distance

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3 years ago

In the middle of a turn, you should _____________________. A. brake hard to maintain complete control over your vehicle B. accel

Elena L [17]

Answer:

D. Downshift to allow you to turn more sharply

3 0

3 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago