Omplete the following program: [0.5 X 4 = 2]

Give me uses of a grinding machine in agriculture.

Tju [1.3M]

Answer:

The grinding machine is used for roughing and finishing flat, cylindrical, and conical surfaces; finishing internal cylinders or bores; forming and sharpening cutting tools; snagging or removing rough projections from castings and stampings; and cleaning, polishing, and buffing surfaces.

4 0

2 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago

A teenage brain is already fully developed to enable us to manage risks effectively.

Kisachek [45]

Answer:

false

Explanation:

the brain is only really fully devolved by age 26

3 0

2 years ago

Read 2 more answers

Can I get an answer to this question please

crimeas [40]

Answer:

(i) 12 V in series with 18 Ω.

(ii) 0.4 A; 1.92 W

(iii) 1,152 J

(iv) 18Ω — maximum power transfer theorem

Explanation:

As seen by the load, the equivalent source impedance is ...

10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

(36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

The load current is ...

(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

(600 s)(1.92 J/s) = 1,152 J

__

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

7 0

2 years ago

The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface

slamgirl [31]

Answer:

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2 ⇒X=4 ft

A=8 x 10=80 $ft^2$

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

4 0

3 years ago