In order for combustion to occur what needs to happen

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is A. 101.325 kPa. B. 651.605 kPa. C. 448.955 kPa. D. 277.28

guajiro [1.7K]

Answer:

Option C is the correct answer.

Explanation:

Absolute pressure is sum of gauge pressure and atmospheric pressure.

That is

$P_{abs}=P_{gauge}+P_{atm}$

We have

$P_{abs}=550.280 kPa\\\\P_{atm}=1atm=101325Pa=101.325kPa$

Substituting

$P_{abs}=P_{gauge}+P_{atm}\\\\550.280=P_{gauge}+101.325\\\\P_{gauge}=448.955kPa$

Option C is the correct answer.

7 0

3 years ago

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm

lina2011 [118]

Answer:

Difference in height = 7.5 cm

Explanation:

We are given;.

Height of ethyl alcohol;h2 = 20 cm = 0.2 m

Density of glycerin: ρ1 = 1260 kg/m³

Density of ethyl alcohol; ρ2 = 790 kg/m³

To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.

Thus;

P1 = P2

Formula for pressure is; P = ρgh

Thus;

ρ1 × g × h1 = ρ2 × g × h2

g will cancel out to give;

ρ1 × h1 = ρ2× h2

Making h1 the subject, we have;

h1 = (ρ2× h2)/ρ1

h1 = (790 × 0.2)/1260

h1 = 0.125 m

Difference in height will be;

Δh = h2 - h1

Δh = 0.2 - 0.125

Δh = 0.075 m = 7.5 cm

4 0

3 years ago

A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s

topjm [15]

Answer:

$\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})$

Explanation:

The electrostatic potential energy is given by the following formula

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Now, we will apply this formula to both cases:

$U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}$

So, the change in the potential energy is

$\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})$

7 0

3 years ago

A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni

irinina [24]

The magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

The given parameters;

length of the solenoid, L = 91 cm = 0.91 m
radius of the solenoid, r = 1.5 cm = 0.015 m
number of turns of the solenoid, N = 1300
current in the solenoid, I = 3.6 A

The magnitude of the magnetic field inside the solenoid is calculated as;

$B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\$

where;

$\mu_o$ is the permeability of frees space = 4π x 10⁻⁷ T.m/A

$B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T$

Thus, the magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

Learn more here:brainly.com/question/17137684

7 0

2 years ago

I have 2 questions

fenix001 [56]

1. your ear drums feel like they pop when you go deep enough under water. kinda feels like bricks. and it hurts if you stay under to long.

2. this could be a reaction from fluid pressure yes. can could be avoided by getting wax ear plugs however even with ear plugs going deep enough under water can still make your ears "pop" due to pressure but at high surfaces it would avoid that.

Please rate me brainliest?

8 0

3 years ago