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Murljashka [212]
4 years ago
10

In Visual Basic/Visual Studio, characteristics of controls, such as the Name of the control, or the Text displayed on the contro

l, are called
Engineering
1 answer:
Arisa [49]4 years ago
8 0

Answer:

Properties

Explanation:

Properties in Visual Basic/Visual Studio consist of the features of the control and these features may include but not limited to he name of the control or even the text being displayed.

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A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates wi
____ [38]

Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

V_{max}=\omega A

now by putting the values

V_{max}=\omega A

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

6 0
4 years ago
Your driver license will be _____ if you race another driver on a public road, commit a felony using a motor vehicle, or are fou
ser-zykov [4K]

Answer:

Revoked

Explanation:

Your driver license will be revoked if you race another driver on a public road, commit a felony using a motor vehicle, or are found guilty of reckless driving three times in one year.

4 0
2 years ago
While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic
makvit [3.9K]

Answer:

Given data:\\While walking across campus one windy day\\Frontal area, \(A=0.3 m ^{2}\)\\Wind speed \(V=24 Km / hr\)\\The drag coefficient \(C_{D, b}=1.2\)\\The combined mass \(m=75 kg\)\\Umbrella diameter, \(D=1.22 m\)\\Velocity of wind \(V=24 \frac{ km }{ hr }\)\\The rolling resistance \(C_{R}=0.75 \%\)

Solution:

Note: Refer the diagram

Basic equation:\\'s law of motion: \(\sum F_{x}=m a_{x}\)\\Lift coefficient, \(C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}\)\\Drag coefficient, \(C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}\)

From force balance equation:\\\(\sum F_{x}=F_{D}-F_{R}=0\)\\But \(F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}\)\(F_{R}=C_{R} m g\)\\Area of the Umbrella \(A_{u}=\frac{\pi D_{u}^{2}}{4}\)\(A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}\)\(A_{v}=1.17 m ^{2}\)

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), C_{D, x}=1.42

Substituting all parameters,

\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}

Then,

\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}

And the equation becomes,

\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}

Thus the floyds travels at 68.3^{\circ}wind speed.

7 0
4 years ago
Match the benefit of full synthetic oil with engine conditions: All engines
mart [117]

Answer:

1. All engines  exhibit wear over time ⇒ <em>Synthetic Improves engine protection by resisting oil breakdown.</em>

Synthetic oil does not breakdown so easily which means that it protects the engine more and protects it from wearing.

2. Engines are cold at start-up and not while running ⇒ <em> Synthetic provides maximum protection in extreme hot and cold temperature conditions.</em>

By providing protection for the engine during cold and hot conditions, the engine will not be too cold when the car is started up.

3.  If oil is thicker, engines lose power and efficiency ⇒ <em>Synthetic has greater resistance to oil thickening to maintain engine efficiency.</em>

Part of the characteristics of synthetic oil is that it does not get as thick as regular oil which means that the adverse effects of thick oil are spared on the engine.

7 0
3 years ago
A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. D
zhuklara [117]

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

d = \sqrt{4*P / (\pi*\sigma)}

d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm

7 0
3 years ago
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