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elena55 [62]
2 years ago
11

3.8 LAB - Select lesson schedule with multiple joins

Engineering
1 answer:
dem82 [27]2 years ago
4 0

Answer:

The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.

Explanation:

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What is the probability that Tina will NOT wear a white t-shirt on the first day of her trip?
katrin2010 [14]

Answer:

4/5

Explanation:

She is not wearing white t-shirt on the first day so she is wearing the other 4 t-shirt

4 0
3 years ago
Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston–cylinder device, as shown in Figure, with an initia
Rashid [163]

The heat transferred to and the work produced by the steam during this process  is 13781.618 kJ/kg

<h3>​How to calcultae the heat?</h3>

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

= 1/2 x ( 75 + 225) x (5 - 2)

W = 450 KJ

From the First Law of Thermodynamics, Q = U + W

So, Heat Transfer = Change in Internal Energy + Work Done

= 13331.618 + 450

Q = 13781.618 kJ/kg

Learn more about heat on:

brainly.com/question/13439286

#SP1

6 0
2 years ago
As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de
katen-ka-za [31]

yes it will

Explanation:

5 0
4 years ago
The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb&gt;ft. Determine t
irina1246 [14]

Answer:

M = 281.25 lb*ft

Explanation:

Given

W<em>man</em> = 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

L = 15.00 m

M<em>max</em> = ?

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0  (+↑)     ⇒    q'*L - W - q*L = 0

⇒       q' = (W + q*L) / L

⇒       q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒       q' = 13 lb/ft   (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10   (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x   (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x²  (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

then

M(7.5) = 5(7.5)² = 281.25 lb*ft

8 0
3 years ago
A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
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