s it possible to simulate the impact of a bullet made out of lead onto a steel plate in a static study in solidworks simulation?

Answer:

μb = 0.096

μc  = 0.073

Explanation:

member AB:

-800( 4/3 ) + Nb (2) = 0

Nb (2) = 3200/3

Nb = 533.3N

Post BC:

summation of force along the y axis=0

Nc + Nb + 150(3/5 ) -50(9.81)=0

Nc + 533.3 + 150(3/5 ) -50(9.81)=0

Nc = 933.83 N

Also (-4/5)(150)(3) + Fb(0.7)= 0

Fb = (4/5)(150)(3)/0.7 = 51.429 N

Likewise alog the x axis,

4/5(150) - Fc -Fb = 0

4/5(150) - Fc -51.429 = 0

Fc = 4/5(150)  -51.429 =68.571 N

μb = Fb/Nb = 51.429/533.3  = 0.096

μc = Fc/Nc = 68.571 / 933.83 = 0.073

A common application that is used for the rod and tube type of control is in a duct-mounted limit control.

A control system can be defined as a type of input and output system that is designed and developed to typically maintain stable (constant) conditions and processes, usually without excessive human interference.

In Control engineering, a duct-mounted limit control is a control system that is commonly used for the operation of the rod and tube type of control.

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Altitude has a direct effect on the specific endurance for a jet aircraft. As altitude increases, the specific endurance of the aircraft decreases.

The specific endurance of an aircraft refers to the amount of time an aircraft can remain in the air on a given amount of fuel. At higher altitudes, the air is thinner and there is less oxygen, which causes the engines to work harder to maintain the same level of performance. This results in a decrease in the specific endurance of the aircraft. Therefore, to maintain the same specific endurance, the aircraft needs to carry more fuel, which makes it heavier and reduces its performance.

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Answer:

D would be correct because it maximizes it.

The ways to remove a core plug are:

Core plugs are known to be called freeze plugs as they are often used to fill the sand casting core holes seen inside water-cooled internal combustion engines.

Note that to do the above, one needs:

The ways to remove a core plug are:

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This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:  

the increase in the overturning moment is 10.1 × 10⁶ lb.ft/ft

Option c) 10.1 × 10⁶ lb.ft/ft id the correct answer

Explanation:

 Given the data in the question and illustrated in the image below;

Hydrostatic force on Dam ( per unit length )

P = \(\frac{1}{2}\) rH.H = \(\frac{1}{2}\)rH² lb/ft

Now, Overturning moment;

M = R × P = H/3 × \(\frac{1}{2}\)rH²  = \(\frac{1}{6}\)rH³ lb.ft/ft

so when H = 175 ft

M₁ =  \(\frac{1}{6}\)r × ( 175 )³  =  \(\frac{1}{6}\) × 62.4 × ( 175 )³

M₁ = 55737500

Also, When H = 185 ft

M₂ =  \(\frac{1}{6}\)r × ( 185 )³  =  \(\frac{1}{6}\) × 62.4 × ( 185 )³

M₂ = 65848900

∴ Increase in overturning moment will be;

⇒ M₂ - M₁ = 65848900 - 55737500

⇒ 10111400 ≈ 10.1 × 10⁶ lb.ft/ft

Therefore, the increase in the overturning moment is 10.1 × 10⁶ lb.ft/ft

Option c) 10.1 × 10⁶ lb.ft/ft id the correct answer

Warfare strategizing is an intellectually rigorous activity that requires careful planning and analysis.

It involves evaluating different options and potential outcomes, and making informed decisions based on the available information. This process can be challenging and complex, as it requires a deep understanding of military tactics and a keen sense of judgment. Despite the mental demands of strategizing, it is a critical component of warfare and plays a crucial role in achieving victory. Warfare strategizing is an activity best described as (a) intellectually rigorous. It requires a deep understanding of military tactics, extensive knowledge of the enemy's strengths and weaknesses, and the ability to make strategic decisions under pressure. While it may also be spiritually demanding, physically tiring, or intuitive, the primary characteristic of warfare strategizing is its intellectual rigor.

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Answer:

a. True

Explanation:

Answer: Shielding gas solenoid

Explanation:

Answer:

R = 20Ω

L = 0.1 H

C = 1 × 10⁻⁵ F

Explanation:

Given the data in the question;

Vs = 10∠30°V   { peak value }

V"s\(_{rms\) = 10/√2 ∠30° V

resonance freq w₀ = 10³ rad/s

Average Power at resonance Power\(_{avg\)  = 2.5 W

Q = 5

values of R, L, and C = ?

We know that;

Power\(_{avg\) = |V"s\(_{rms\)|² / R

{ resonance circuit is purely resistive }

we substitute

2.5 = (10/√2)² × 1/R

2.5 = 50 × 1/R

R = 50 / 2.5

R = 20Ω

We also know that;

Q = w₀L / R

we substitute

5 = ( 10³ × L ) / 20

5 × 20 = 10³ × L

100 = 10³ × L

L = 100 / 10³

L = 0.1 H

Also;

w₀ = 1 / √LC

square both side

w₀² = 1 / LC

w₀²LC = 1

C = 1 / w₀²L

we substitute

C = 1 / [ (10³)² × 0.1 ]

C = 1 / [ 1000000 × 0.1 ]

C = 1 / [ 100000 ]

C = 0.00001 ≈ 1 × 10⁻⁵ F

Therefore;

R = 20Ω

L = 0.1 H

C = 1 × 10⁻⁵ F

The cubic inch space that is required inside a box for 4 #6 XHHN current carrying conductors will be D. 8 cubic inches.

It should be noted that a conductor simply means a substance or material that simply allows electricity to pass through it.

From the information given, the cubic inch space that is required inside a box for 4 #6 XHHN current carrying conductors is yo be computed.

This will be:

= (4 × 6)/3

= 24/3

= 8

In conclusion, the cubic inch space that is required inside a box for 4 #6 XHHN current carrying conductors will be 8 cubic inches.

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Answer:D

Explanation:

All the options given regarding the hazard communication standard are correct except option D "prepare and post a list of employees who handle hazardous and toxic substances in the workplace".

The main requirements of employers that use hazardous chemicals are:

• ensure that chemicals are properly labelled.

• provide safety data sheets.

• train employees.

• create a written hazard communication program.

It should be noted that preparing and posting a list of employees who handle hazardous and toxic substances in the workplace is incorrect.

Therefore, the correct option is D.

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Answer:

2.9237 ft/s

Explanation:

Given the data in the question;

A₁ = 9-inch × 9-inch = 81 in² = 81 / 144 = 0.5625 ft²

V₁ = 6.1 ft/s

A₂ = 13 in × 13 in = 169 in² = 1.17361 ft²

v₂ = ?

using the the equation if continuity

( Rate of volumetric flow is constant )

A₁V₁ = A₂V₂

we substitute

0.5625 ft² × 6.1 ft/s = 1.17361 ft² × V₂

3.43125 ft³/s = 1.17361 ft² × V₂  

V₂  = 3.43125 ft³/s  / 1.17361 ft²

V₂ = 2.9237 ft/s

Therefore, the speed of the air flowing into the room is 2.9237 ft/s

At 1.2mpa pressure and 50c

What is pressure?
By pressing a knife against some fruit, one can see a straightforward illustration of pressure. The surface won't be cut if you press the flat part of the knife against the fruit. The force is dispersed over a wide area (low pressure).

a)Mass flow rate of the refrigerant
Therefore h1= condenser inlet enthalpy =278.28KJ/Kg
saturation temperature at 1.2mpa is 46.29C
Therefore the temperature of the condenser

T2 = 46.29C - 5
T2 = 41.29C

Now,
d)power consumed by compressor W = 3.3KW
Q4 = QL + w = Q4
QL = mR(h1-h2)-W
= 0.0498 x (278.26 - 110.19)-3.3
=5.074KW
Hence refrigerator load is 5.74Kg

(COP)r = 238/53
(Cop) = 4.490

Therefore the above values are the (a) mass flow rate of the refrigerant, b) the refrigerant load, c) the cop, and d) the minimum power input to the compressor for the same refrigeration load.

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Answer:

Diameter of pipe is 0.535 ft

Explanation:

see attachment, its works out 1st half

The process of data processing by a computer involves several steps that take place in a specific order. The first step is inputting the data into the computer system. This can be done through various devices such as keyboards, scanners, or touchscreens. Once the data is entered into the computer, it is stored in the memory for further processing.

The second step is processing the data. This involves the use of the central processing unit (CPU) which performs mathematical and logical operations on the data. The CPU retrieves the instructions from the memory and executes them, which results in the desired output.

The third step is storing the output data. Once the CPU has processed the data, the result is stored in the memory. The output can be in various forms such as text, images, or sound.

The fourth step is displaying the output. The output is displayed on the screen, printer, or other output devices depending on the nature of the data and the desired format.

Finally, the last step is transmitting the output. If the output is required to be sent to another system or device, it is transmitted using various modes such as email, internet, or other networks.

Overall, these steps are followed in a sequential order to ensure efficient and accurate data processing by the computer. The speed and accuracy of the process depend on the hardware and software components of the computer system, as well as the complexity of the data being processed.

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An advanced distance-vector routing protocol is EIGRP.

Among the given options, EIGRP (Enhanced Interior Gateway Routing Protocol) is considered an advanced distance-vector routing protocol. EIGRP combines features of both distance-vector and link-state protocols, providing fast convergence and efficient use of network resources. It uses a metric based on bandwidth, delay, reliability, and load to calculate the best path for routing traffic. EIGRP also supports features like load balancing, route summarization, and route authentication. It is primarily used in larger networks where scalability and fast convergence are important factors. OSPF (Open Shortest Path First) and BGP (Border Gateway Protocol) are examples of link-state and path-vector routing protocols, respectively.

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Answer:

Fourier coefficient, Ck:

\({ \rm{ c_{k}} = \int ^{ \infin} _{ - \infin } \{x(t) {e}^{jk \omega _{0} t} \} }\)

Fourier transform:

\({ \rm{x( \omega) = \int ^{ \infin} _{ - \infin} } \{x(t) {e}^{j \omega} t} \}\)

Answer:

  5

Explanation:

One pulse is required for each bit to be moved.

5 pulses are required.

Answer: true

Explanation:

If lift and duration remain constant and the lobe center angle decreases, Option A: the valve overlap decreases.

Valve overlap refers to the duration when both the intake and exhaust valves are simultaneously open in the course of the engine's cycle.

The timing of valve opening and closing with respect to one another is dictated by the angle of the center of the lobe. If the angle of the center of the lobe diminishes, the opening and shutting of the valves will occur more proximately, leading to a reduction in the duration of valve overlap.

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These are some tips for refining search results:A) Try different keywords: If you are not getting the desired results, try using different keywords that are related to your search.

Use synonyms or alternate phrasing to broaden your search and increase the chances of finding relevant results.Try more general keywords: If you are using very specific or niche keywords, try using more general keywords that are related to your search. This can help to uncover results that may have been missed using specific search terms.Try fewer keywords: If your search is too specific and you are not getting any results, try using fewer keywords. This can help to broaden your search and increase the chances of finding relevant results.

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Answer: Attached below is the missing detail and Mohr's circle.

i) б1 =  9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii)  -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

      = -20 / 0.785 * 5^2  = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

  = (16 * 250 ) / π * 5^3  = 010.19 Ksi

Calculate the Principal stress, maximum in-plane shear stress and average normal stress

Using Mohr's circle ( attached below )

i) principal stresses:

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

ii) maximum in-plane shear stress

ζ  = radius of Mohr's circle

   = 5.1 cm = 10.2 Ksi   ( Given that ; 1 cm = 2Ksi )

iii) average normal stress

 = 9.6 + ( - 10.7 ) / 2

  = -0.51Ksi

To determine the peak stress on the outer surface of a 50% fibre composite vaulting pole that has a diameter of about 50 mm and is being bent to a radius of curvature, R, of about 0.5 m, we can use the following formula:

σ = (M/Z) * y

where,σ = stress

M = bending moment

Z = section modulus

y = distance from the neutral axis

To find M, we can use the following formula:

M = (Em/Ef) * (ρc/ρf) * (π/4) * d^3 * (1/R)where,

Em = modulus of elasticity of matrixEf = modulus of elasticity of fibre

ρc = density of the matrixρf = density of the fibreπ = 3.14d = diameter of the pole

R = radius of curvatureNow, substituting the given values:

M = (3/90) * (0.5/1.5) * (π/4) * (0.05)^3 * (1/0.5)M

= 0.000216 Nm

To find Z, we can use the following formula:

Z = (π/32) * d^4 * (ρf/ρc) * (Ef/Em) * (1 + (3h/d)^2)

where,h = height of the pole

Now, substituting the given values:

Z = (π/32) * (0.05)^4 * (0.5/1.5) * (90/3) * (1 + (3(0.5)/0.05)^2)Z

= 0.0001586 m^3To find y, we can use the following formula:

y = (d/2) * (ρf/ρc)where,ρc = density of the matrix

ρf = density of the fibre

Now, substituting the given values:y = (0.05/2) * (0.5/1.5)y = 0.00833 m

Substituting the values of M, Z, and y in the formula for stress, we get:σ = (M/Z) * yσ = (0.000216/0.0001586) * 0.00833σ = 11.77 MPa

Therefore, the peak stress on the outer surface of a 50% fibre composite vaulting pole with a diameter of about 50 mm and being bent to a radius of curvature, R, of about 0.5 m is 11.77 MPa.

Composite materials are known for their remarkable strength and durability, making them ideal for use in a variety of applications. One of these applications is in the production of vaulting poles used in pole vaulting, a track and field event that involves athletes using a long, flexible pole to clear a bar that is suspended above the ground. The poles used in this event are typically made of a composite material that consists of a matrix material, such as epoxy resin, and a reinforcing material, such as carbon fibre or fibreglass. These materials are chosen for their high strength-to-weight ratio, which allows the poles to be both strong and flexible. The strength of the pole is determined by its modulus of elasticity, which is a measure of its ability to resist deformation under load. The modulus of elasticity of the matrix material, Em, and the reinforcing material, Ef, are key factors in determining the strength of the pole. The density of these materials is also important, as it affects the weight of the pole. The peak stress on the outer surface of a pole is an important factor to consider when designing the pole, as it determines the maximum load that the pole can handle before it fails. In order to determine the peak stress on the outer surface of a 50% fibre composite vaulting pole with a diameter of about 50 mm and being bent to a radius of curvature, R, of about 0.5 m, we can use the formula σ = (M/Z) * y, where M is the bending moment, Z is the section modulus, and y is the distance from the neutral axis. By using the appropriate values for Em, Ef, density, diameter, and radius of curvature, we can calculate the peak stress on the outer surface of the pole.

In conclusion, the peak stress on the outer surface of a 50% fibre composite vaulting pole with a diameter of about 50 mm and being bent to a radius of curvature, R, of about 0.5 m, is 11.77 MPa. The strength of the pole is determined by its modulus of elasticity, which is a measure of its ability to resist deformation under load. The modulus of elasticity of the matrix material, Em, and the reinforcing material, Ef, are key factors in determining the strength of the pole. The density of these materials is also important, as it affects the weight of the pole. The peak stress on the outer surface of a pole is an important factor to consider when designing the pole, as it determines the maximum load that the pole can handle before it fails. Overall, composite vaulting poles are an excellent example of how composite materials can be used to create strong and durable structures that are ideal for a wide range of applications.

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Answer:

b remains the same

Explanation:

voltage and amps have no connection

the electricity used to rin your clothes dryer is normally 220V 18-24A

the voltage in your car's battery is usually 12V 20-30A

see they are approximately the same amperage but very different voltage

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

\(\sigma = \frac{F}{A}\\\\\)

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5\(\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2\)

Therefore,

\(\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips\)

Now, we determine the strain:

\(strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}\)

Now, the modulus of elasticity (E) is given as:

\(E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}\)

E = 8.83 kips

Answer: up to seven individual 6-gauge wires,

Explanation: while 1-inch schedule 40 rigid PVC can fit six. This is only true for 6 gauge wire.

The answer depends on the type of conduit, the length of the conduit run, the type of insulation on the wire, and the installation requirements. However, as a general rule of thumb, you can safely run three to four 6 AWG wires through a 1-inch conduit.

A conduit is a metal or plastic pipe that encloses and protects electrical wires. In electrical installations, the size of the conduit and the number of wires that can be run through it depend on various factors such as the wire gauge and insulation, conduit type, and installation requirements. The American Wire Gauge (AWG) is a standard system used to specify the diameter of electrical wires. The higher the AWG number, the smaller the wire diameter. The AWG size also determines the amount of current that can be carried by the wire. It is important to note that cramming too many wires into a conduit can cause overheating, which can lead to a fire hazard. Therefore, it is recommended to consult the National Electrical Code (NEC) or a qualified electrician for specific guidance on wire and conduit sizing for your particular installation.

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