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3 years ago

10. To cut 1/4" (6 mm) thick mild steel at a rate of 40 inches per minute, the current would be set to

Engineering

1 answer:

pishuonlain [190]3 years ago

5 0

Answer: Idk but try downloading more apps it’s a lot easier to have more than one

Explanation:

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Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5s

zlopas [31]

Answer:

4m/s

Explanation:

We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted

Mathematically

$Power_{motor} } =\frac{Energy }{time}$ \

We also know that Power = force x velocity ..................(i)

The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load

=> power = mg x Velocity........(ii)

While hoisting the load at at constant speed only the potential energy of the mass increases

Thus Potential energy = Mass x g x H...................(iii)

where

g = accleration due to gravity (9.81m/s2)

H = Height to which the load is hoisted

Equating equations (ii) and (iii) we get

m x g x v = $\frac{mgh}{t}$

thus we get v = H/t

Applying values we get

v = 6/1.5 = 4m/s

5 0

3 years ago

Klio2033 [76]

The C++ code that would draw all the iterations in the selection sort process on the array is given below:

#include <stdio.h>

#include <stdlib.h>

int main() {

int i, temp1, temp2;

int string2[16] = { 0, 4, 2, 5, 1, 5, 6, 2, 6, 89, 21, 32, 31, 5, 32, 12 };

_Bool check = 1;

while (check) {

temp1 = string2[i];

temp2 = string2[i + 1];

if (temp1 < temp2) {

string2[i + 1] = temp1;

string2[i] = temp2;

i = 0;

} else {

i++;

if (i = 15) {

check = !check;

}

return 0;

}

Read more about C++ programming here:

brainly.com/question/20339175

#SPJ1

5 0

1 year ago

F.R.E.E P.O.I.N.T.S F.R.E.E PO.I.N.T.S

grigory [225]

Answer:

yayyy thank you

lol

Explanation:

7 0

2 years ago

Read 2 more answers

A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If t

maksim [4K]

Answer:

A) centerline velocity = 1.894 m/s

B) flow rate = 7.44 x 10^(-3) m³/s

Explanation:

A) The flow velocity intensity for the input radial coordinate "r" is given by;

U(r) = (Δp•D²/16μL) [1 - (2r/D)²]

Velocity at the centre of the tube can be expressed as;

V_c = (Δp•D²/16μL)

Thus,

U(r) = (V_c)[1 - (2r/D)²]

From question, diameter = 0.1m,thus radius (r) = 0.1/2 = 0.05m

But we are to find the velocity at the centre of the tube, thus;

We will use the radius across the horizontal distance which will be;

0.05 - 0.012 = 0.038m

Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.

Thus;

U(r) = (V_c)[1 - (2r/D)²]

0.8 = (V_c)[1 - {(2 * 0.038)/0.1}²]

0.8 = (V_c)[1 - (0.76)²]

V_c = 0.8/0.4224 = 1.894 m/s

B) flow rate is given by;

ΔV = Average Velocity x Area

Now, average velocity = V_c/2

Thus, average velocity = 1.894/2 = 0.947 m/s

Area(A) = πr² = π x 0.05² = 0.007854 m²

So, flow rate = 0.947 x 0.007854 = 7.44 x 10^(-3) m³/s

4 0

3 years ago

: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the

Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

$\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}$

$V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s$

$\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}$

taking f = 0.0185

at Z₁ = Z₂

$\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}$

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P= 1.13\ psi$

c) When flow is downhill z₂-z₁ = -2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P=-0.601\ psi$

7 0

3 years ago