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AysviL [449]
4 years ago
6

In which of the fire plume structure zones do the gases accelerate upward?

Physics
1 answer:
I am Lyosha [343]4 years ago
3 0
The zone that gases always accelerate upward is the Luminous flame zone. The fire plume is the column of hot gases, flames and smoke rising above a fire. Gases accelerate upward toward the always luminous flame zone. The luminous flame height is the distance between the base of a flame and the point at which the plume is luminous half the time and transparent half the time.
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A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250
Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30

coefficient of static friction \mu =0.35

coefficient of kinetic friction \mu _k=0.25

distance traveled d=77.3 cm

spring constant k=35 N/m

work done by gravity+work done by friction=Energy stored in Spring

mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}

mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}

0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}

x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}

x=0.234 m

x=23.4 cm

6 0
3 years ago
If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy pf 392,000 J and a kinetic ene
Viefleur [7K]
I just woke him crying crying laughing crying and laughing I’m so mad he got me blocked him lol I got a hold on her phone the answer is 0 m
5 0
3 years ago
Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

8 0
3 years ago
A 3.00 x 10^2-W electric immersion heater is
andre [41]

Answer

t = 367.77 s = 6.13 min

Explanation:

According to the law of conservation of energy:

Heat\ Supplied\ By \ Heater = Heat\ Absorbed\ by\ Glass + Heat\ Absorbed\ by\ Water\\Pt = m_gC_g\Delta T_g + m_wC_w\Delta T_w\\

where,

P = Electric Power of Heater =  300 W

t = time required = ?

m_g = mass of glass = 300 g = 0.3 kg

m_w = mass of water = 250 g = 0.25 kg

C_g = speicific heat of glass = 840 J/kg.°C

C_w =  specific heatof water = 4184 J/kg.°C

ΔT_g = ΔT_w = Change in Temperature of Glass and water = 100°C - 15°C

ΔT_g = ΔT_w = 85°C

Therefore,

(300\ W)(t) = (0.3\ kg)(840\ J/kg.^oC)(85^oC)+(0.25\ kg)(4184\ J/kg.^oC)(85^oC)\\

<u>t = 367.77 s = 6.13 min</u>

8 0
3 years ago
Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p
Temka [501]

Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT...............................  Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

6 0
3 years ago
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