Identify the branch of study each student should pursue.

You installed a new 40 gallon water heater with a 54,000 BTUh burner. The underground water temperature coming into the house is

kipiarov [429]

14256000. Kanjiuijhgg

5 0

3 years ago

Read 2 more answers

What properties should the head of a carpenter’s hammer possess? How would you manufacture a hammer head?

BabaBlast [244]

Properties of Carpenter's hammer possess

Explanation:

1.The head of a carpenter's hammer should possess the impact resistance, so that the chips do not peel off the striking face while working.

2.The hammer head should also be very hard, so that it does not deform while driving or eradicate any nails in wood.

3.Carpenter's hammer is used to impact smaller areas of an object.It can drive nails in the wood,can crush the rock and shape the metal.It is not suitable for heavy work.

How hammer head is manufactured :

1.Hammer head is produced by metal forging process.

2.In this process metal is heated and this molten metal is placed in the cavities said to be dies.

3.One die is fixed and another die is movable.Ram forces the two dies under the forces which gives the metal desired shape.

4.The third process is repeated for several times.

5 0

3 years ago

A water tank filled with solar-heated water at 40°C is to be used for showers in a field using gravity-driven flow.

Alenkasestr [34]

Answer:

yes sir

Explanation:

4 0

3 years ago

Air fl ows isentropically through a duct. At section 1, the pressure and temperature are 250 kPa and 1258C, and the velocity is

abruzzese [7]

The correct temperature is 125°C

Answer:

A) M_a1 = 0.5

B) T2 = 232.17 K

C) V2 = 611 m/s

D) m' = 187 kg/s

Explanation:

We are given;

Pressure; P1 = 250 kPa

Temperature; T1 = 125°C = 398 K

Speed; v1 = 200 m/s

Area; A2 = 0.25 m²

M_a2 = 2

A) Formula for M_a1 is given by;

M_a1 = v/a1

Where;

v is speed

a1 = √kRT

k is specific heat capacity ratio of air = 1.4

R is a gas constant with a value of R = 287 J/kg·K

T is temperature

Thus;

M_a1 = 200/√(1.4 × 287 × 398)

M_a1 = 200/399.895

M_a1 = 0.5

B) To find T2, let's first find the Stagnation pressure T0

Thus;

T0/T1 = 1 + ((k - 1)/2) × (M_a1)²

T0 = T1(1 + ((k - 1)/2) × (M_a1)²)

T0 = 398(1 + ((1.4 - 1)/2) × (0.5)²)

T0 = 398(1 + (0.2 × 0.5²))

T0 = 398 × 1.05

T0 = 417.9 K

Now,similarly;

T0/T2 = 1 + ((k - 1)/2) × (M_a2)²

T2 = T0/[(1 + ((k - 1)/2) × (M_a2)²)]

T2 = 417.9/(1 + (0.2 × 2²))

T2 = 417.9/1.8

T2 = 232.17 K

C) V2 is gotten from the formula;

T0 = T2 + (V2)²/(2C_p)

Cp of air = 1005 J/Kg.K

Thus;

V2 = √(2C_p)[T0 - T2]

V2 = √((2 × 1005) × (417.9 - 232.17))

V2 =√373317.3

V2 = 611 m/s

D) mass flow is given by the formula;

m' = ρA2•V2

Where;

ρ is Density of air with an average value of 1.225 kg/m³

m' = 1.225 × 0.25 × 611

m' = 187 kg/s

6 0

3 years ago

**Please Help, ASAP**

Novay_Z [31]

Explanation:

We need to rearrange the following formula for the values given in parenthesis.

(1) x+xy = y, (x)

taking x common in LHS,

x(1+y)=y

$x=\dfrac{y}{1+y}$

(2) x+y = xy, (x)

Subrtacting both sides by xy.

x+y-xy = xy-xy

x+y-xy = 0

x-xy=-y

x(1-y)=-y

$x=\dfrac{-y}{1-y}$

(3) x = y+xy, (x)

Subrating both sides by xy

x-xy = y+xy-xy

x(1-y)=y

$x=\dfrac{y}{1-y}$

(4) E = (1/2)mv^2-(1/2)mu^2, (u)

Subtracting both sides by (1/2)mv^2

E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2

E-(1/2)mv^2 =-(1/2)mu^2

So,

$2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}$

(5) (x^2/a^2)-(y^2/b^2) = 1, (y)

$\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}$

(6) ay^2 = x^3, (y)

$y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}$

Hence, this is the required solution.

3 0

3 years ago