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2 years ago

How many parts of oxygen do you need for each part of acetylene for a neutral flame?

Physics

1 answer:

chubhunter [2.5K]2 years ago

8 0

Two parts have a good day

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A jet taxiing down the runway receives word that it must return to the gate. The jet is traveling 37.6 m/s when the pilot receiv

IrinaVladis [17]

Answer:

7 m/s^2

Explanation:

Given that the jet is traveling 37.6 m/s when the pilot receives the message.

And it takes the pilot 5.37 s to bring the plane to a halt.

Acceleration of the plane can be calculated by using first equation of motion

V = U - at

Since the plane is going to stop, the final velocity V = zero.

And the acceleration will be negative

Substitute all the parameters into the formula

0 = 37.6 - 5.37a

5.37a = 37.6

Make a the subject of formula

a = 37.6 / 5.37

a = 7.0 m/s^2

Therefore, the acceleration of the plane to bring the plane to a halt is 7 m/s^2

5 0

2 years ago

20.

sveticcg [70]

Answer:

I believe the answer is A and D.

I am unsure of C.

4 0

2 years ago

Read 2 more answers

A bug is sitting on the edge of a rotating disk. At what angular velocity will the bug slide off the disk if its radius is 0.241

Ivahew [28]

Answer:

ω = 3.61 rad/sec

Explanation:

Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.

μmg = mv^2/r = mω^2r

Thus;

μg = ω^2r

ω^2 = μg/r

ω = √(μg/r)

ω = √(0.321 * 9.8)/0.241

ω = √(13.05)

= 3.61 rad/sec

3 0

2 years ago

What is newton's first low of motion?

sergey [27]

Newton's first law of motion is that an object in motion will tend to stay in motion unless an external force acts upon it.

8 0

2 years ago

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calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

2 years ago