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Ket [755]
2 years ago
9

Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional

courses in specialized areas. What are these courses in specialized areas called? A. technical certificate courses B. undergraduate certificate courses C. master’s degree courses D. online courses
Engineering
1 answer:
Anit [1.1K]2 years ago
4 0

Answer:

i dont know

Explanation:

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A glass tube is inserted into a flowing stream of water with one opening directed upstream and the other end vertical. If the wa
Furkat [3]

Answer:

h=0.46m

Explanation:

From the question we are told that:

Velocity of water V=3m/s

Height=?

Generally, the equation for Water Velocity is mathematically given by

V=\sqrt{2gh}

Therefore Height h is given as

h=\frac{v}{2g}

h=\frac{3^2}{2*9.81}

h=0.46m

5 0
3 years ago
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str
Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 mm^4

So the shear stress at point  A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

3 0
3 years ago
The atmosphere within a room is at 70 °F dry-bulb temperature, 50 percent degree of saturation, and 14.696 psia pressure. The in
Gre4nikov [31]

Answer:

Given that the temperature of the window is below the dew point it will condensate.

Explanation:

A psychrometric chart (like the one attached) will give you the information needed. This chart is for 14.696 psia.

On the bottom horizontal axes you have the dry-bulb temperature, in this case 70°F, going up from this point you can reach the 50% relative humidity curve (red point on chart), going horizontally from this point to the 100% relative humidity you get the dew point temperature (the point at which moisture will condensate) (blue point on chart). In this case the dew point is 50°C. Given that the temperature of the window is below the dew point it will condensate.

6 0
3 years ago
The tolerance for a geometric can be easily identified by reading the _________.
baherus [9]

Answer:

C

Explanation:

4 0
3 years ago
Read 2 more answers
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
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