What form of joining uses heat to create coalescence of the materials?

Lois is in high school and enjoys studying math and science subjects. She aspires to get into the robotics industry after comple

Inessa [10]

Answer:

d.partivipate in a robotic club

hope helps

mark me please

4 0

3 years ago

A book of the Law was found in the Temple, which was being repaired, during the___

grigory [225]

Answer:

2

Explanation:

5 0

3 years ago

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A 65% efficient turbine receives 2 m^3/s of water from a reservoir. The reservoir water surface is 45 m above the centerline of

laiz [17]

Answer:

$P_{out} = 508.071 kW$

Given:

efficiency of the turbine, $\eta$ = 65% = 0.65

available gross head, $H_{G}$ = 45 m

head loss, $H_{loss}$ = 5 m

Discharge, Q = $2 m^{3}$

Solution:

The nozzle is 100% (say)

Available power at the inlet of the turbine, $P_{inlet}$ is given by:

$P_{inlet} = \rho Qg(H_{G} - H_{loss})$ (1)

where

$\rho$ = density of water = 997 $kg/m^{3}$

acceleration due to gravity, g = $9.8 m^{2}$

Using eqn (1):

$P_{inlet} = 997\times 2\times 9.8(45 - 5) = 781.65 kW$

Also, efficency, $\eta$ is given by:

$\eta = \frac{P_{out}}{P_{inlet}}$

$0.65 = \frac{P_{out}}{781.648\times 1000}$

$P_{out} = 0.65\times 781.648\times 1000 = 508071 W = 508.071 kW$

$P_{out} = 508.071 kW$

3 0

3 years ago

What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-

Hitman42 [59]

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

$\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}$

Where $\sigma_{m}$ is the magnitude of he maximum stress at the tip of the crack, $\sigma_{p}$ is the magnitude of the tensile stress, $l_{c}$ is $1/2$ the length of the internal crack, and $r_{c}$ is the radius of curvature of the crack.

We have:

$r_{c}=1.9*10^{-4} [mm]$

$l_{c}=3.8*10^{-2} [mm]$

$\sigma_{c}=140 [MPa]$

We replace:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}$

We get:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]$

5 0

3 years ago

A 1:50 scale model of a ship is towed at 4.8 km/hr using a force of 9 N. If we assume the same fluid in the model as in the prot

Mashutka [201]

Answer:

A: density and gravity

Explanation:

The Froude Number is defined as a dimensionless parameter that measures the ratio of the force of inertia on an element of fluid to the weight of the fluid element. In simple terms, it's the force of inertia divided by the gravitational force.

Froudes number is usually expressed as;

Fr = v/√(gd)

Where;

Fr = froude number

v = velocity

g = gravitational acceleration = specific weight/density

d = depth of flow

Now, to calculate the corresponding speed and force in the prototype, it means we have to use equal froude number and thus this will mean that it has to be dominated by gravity and density.

4 0

3 years ago