Why must air tanks be drained

An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).

Phoenix [80]

Answer:

The right choice would be Option b (2.545).

Explanation:

The given values are:

The aggregate blend will be:

$W_A$ = 55%

$G_ A$ = 2.631

$W_ B$ = 25%

$G_B$ = 2.331

$W_C$ = 20%

$G_C$ = 2.609

Now,

On applying the formula, we get

⇒ $G_{BA}=\frac{W_A+W_B+W_C}{\frac{W_A}{G_A} +\frac{W_B}{G_B} +\frac{W_C}{G_C}}$

On substituting the values, we get

⇒ $=\frac{55+25+20}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}$

⇒ $= \frac{100}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}$

⇒ $=2.545$

8 0

3 years ago

6. What are the potential consequences if a business does not follow regulations? (1 point)

dimaraw [331]

Answer:

all of the above

Explanation:

6 0

3 years ago

Assign deliveryCost with the cost (in dollars) to deliver a piece of baggage weighing baggageWeight. The baggage delivery servic

Sholpan [36]

Answer: y = x * 1dollars - 30dollars

Explanation:

Giving that the delivery cost in dollar is potent for all x > = 50 pounds of wght

Y = (x - 50)*1 dollar + c ...equ 1

Y = delivery cost equation in dollars

x = weigt of baggage for delivery

c = 20dollars = down payment for the first 50 pound weight of baggage

Equ 1 becomes

Y = (x)dollars - 50 dollars + 20 dollars

Y = (x) dollars -30 dollars

4 0

4 years ago

III. During January, at a location in Alaska winds at −27°C can be observed. However, several meters below ground the temperatur

Naya [18.7K]

Answer:

Not possible.

Explanation:

According to second law of thermodynamics, the maximum efficiency any heat engine could achieve is Carnot Efficiency η defined by:

$\eta=1-\frac{T_{cold}}{T_{hot}}$

Where

$T_{hot}$ and $T_{cold}$ are temperature (in Kelvin) of heat source and heatsink respectively

In our case (I will be using K = 273+°C) :

$\eta=1-\frac{-27+273}{14+273}\\=0.1428$

In percentage, this is 14.28% efficiency, which is the maximum theoretical efficiency any heat engine could have while working between -27 and 14 °C temperature. Any claim of more efficient heat engine between these 2 temperature are violates the second law of thermodynamics. Therefore, the claim must be false.

6 0

3 years ago

What are the de Broglie frequencies and wavelengths of (a) an electron accelerated to 50 eV (b) a proton accelerated to 100 eV

DaniilM [7]

Answer:

(a) De-Brogie wavelength is 0.173 nm and frequency is 2.42 x 10^16 Hz

(b) De-Brogie wavelength is 2.875 pm and frequency is 4.8 x 10^16 Hz

Explanation:

(a)

First, we need to find velocity of electron. Since, it is accelerated by electric potential. Therefore,

K.E of electron = (1/2)mv² = (50 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 8 x 10^(-18) J

Mass of electron = m = 9.1 x 10^(-31) kg

Therefore,

v² = [8 x 10^(-18) J](2)/(9.1 x 10^(-31) kg)

v = √1.75 x 10^13

v = 4.2 x 10^6 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(9.1 x 10^(-31) kg)(4.2 x 10^6 m/s)

λ = 0.173 x 10^(-9) m = 0.173 nm

The frequency is given as:

Frequency = f = v/λ

f = (4.2 x 10^6 m/s)/(0.173 x 10^(-9) m)

f = 2.42 x 10^16 Hz

(b)

First, we need to find velocity of proton. Since, it is accelerated by electric potential. Therefore,

K.E of proton = (1/2)mv² = (100 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 1.6 x 10^(-17) J

Mass of proton = m = 1.67 x 10^(-27) kg

Therefore,

v² = [1.6 x 10^(-17) J](2)/(1.67 x 10^(-27) kg)

v = √1.916 x 10^10

v = 1.38 x 10^5 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(1.67 x 10^(-27) kg)(1.38 x 10^5 m/s)

λ = 2.875 x 10^(-12) m = 2.875 pm

The frequency is given as:

Frequency = f = v/λ

f = (1.38 x 10^5 m/s)/(2.875 x 10^(-12) m)

f = 4.8 x 10^16 Hz

6 0

4 years ago

Why must air tanks be drained​

Why must air tanks be drained