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Svetllana [295]
3 years ago
15

I JUST PLAYED A GAME IN 2ND PERSON AND IM CONFEUSED

Engineering
2 answers:
sattari [20]3 years ago
6 0

Why would you be confused

Also which game

Crazy boy [7]3 years ago
3 0

Answer:

cool

Explanation:

You might be interested in
a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8
djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam =  159.07 / 2 = 79.535 MPa  

attached below is the remaining solution

6 0
3 years ago
The __________ developed the national electric code, the national building code, and the national fire prevention code.
Dahasolnce [82]

Main Answer:

<u>The Bureau of Indian standards</u> developed the national electric code, the national building code, and the national fire prevention code.

Sub heading:

explain BIS?

Explanation:

1.BIS-bureau of indian standard is the national standard body of india.

2.BIS is responbility for the harmonious deve;opment of the activities of standardization.marking .

Reference link:

https://brainly.com

Hashtag:

#SPJ4

5 0
2 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
DochEvi [55]

Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

6 0
3 years ago
A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055
Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}

So the mean conductivity

K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )

T_m=\dfrac{160+273+40+273}{2}

T_m=373 K

K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )

K_m=0.112 \frac{W}{mK}

So heat conduction through cylinder

Q=kA\dfrac{\Delta T}{L}

Q=0.112\times \pi \times 0.15^2\times 120

Q=0.95 W/m

4 0
3 years ago
3.3 Equation (2) for VCPP is rather difficult to prove at this time. Take it as a challenge to derive it as you learn increasing
podryga [215]

Answer:

For an RC integrator circuit, the input signal is applied to the resistance with the output taken across the capacitor, then VOUT equals VC. As the capacitor is a frequency dependant element, the amount of charge that is established across the plates is equal to the time domain integral of the current. That is it takes a certain amount of time for the capacitor to fully charge as the capacitor can not charge instantaneously only charge exponentially.

Therefore the capacitor current can be written as:

 

his basic equation above of iC = C(dVc/dt) can also be expressed as the instantaneous rate of change of charge, Q with respect to time giving us the following standard equation of: iC = dQ/dt where the charge Q = C x Vc, that is capacitance times voltage.

The rate at which the capacitor charges (or discharges) is directly proportional to the amount of the resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC integrator circuit is the time interval that equals the product of R and C.

Since capacitance is equal to Q/Vc where electrical charge, Q is the flow of a current (i) over time (t), that is the product of i x t in coulombs, and from Ohms law we know that voltage (V) is equal to i x R, substituting these into the equation for the RC time constant gives:

We have seen here that the RC integrator is basically a series RC low-pass filter circuit which when a step voltage pulse is applied to its input produces an output that is proportional to the integral of its input. This produces a standard equation of: Vo = ∫Vidt where Vi is the signal fed to the integrator and Vo is the integrated output signal.

The integration of the input step function produces an output that resembles a triangular ramp function with an amplitude smaller than that of the original pulse input with the amount of attenuation being determined by the time constant. Thus the shape of the output waveform depends on the relationship between the time constant of the circuit and the frequency (period) of the input pulse.

By connecting two RC integrator circuits together in parallel has the effect of a double integration on the input pulse. The result of this double integration is that the first integrator circuit converts the step voltage pulse into a triangular waveform and the second integrator circuit converts the triangular waveform shape by rounding off the points of the triangular waveform producing a sine wave output waveform with a greatly reduced amplitude.

RC Differentiator

For a passive RC differentiator circuit, the input is connected to a capacitor while the output voltage is taken from across a resistance being the exact opposite to the RC Integrator Circuit.

A passive RC differentiator is nothing more than a capacitance in series with a resistance, that is a frequency dependentTherefore the capacitor current can be written as:

 

 

device which has reactance in series with a fixed resistance (the opposite to an integrator). Just like the integrator circuit, the output voltage depends on the circuits RC time constant and input frequency.

Thus at low input frequencies the reactance, XC of the capacitor is high blocking any d.c. voltage or slowly varying input signals. While at high input frequencies the capacitors reactance is low allowing rapidly varying pulses to pass directly from the input to the output.

This is because the ratio of the capacitive reactance (XC) to resistance (R) is different for different frequencies and the lower the frequency the less output. So for a given time constant, as the frequency of the input pulses increases, the output pulses more and more resemble the input pulses in shape.

We saw this effect in our tutorial about Passive High Pass Filters and if the input signal is a sine wave, an rc differentiator will simply act as a simple high pass filter (HPF) with a cut-off or corner frequency that corresponds to the RC time constant (tau, τ) of the series network.

Thus when fed with a pure sine wave an RC differentiator circuit acts as a simple passive high pass filter due to the standard capacitive reactance formula of XC = 1/(2πƒC).

But a simple RC network can also be configured to perform differentiation of the input signal. We know from previous tutorials that the current through a capacitor is a complex exponential given by: iC = C(dVc/dt). The rate at which the capacitor charges (or discharges) is directly proportional to the amount of resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC differentiator circuit is the time interval that equals the product of R and C. Consider the basic RC series circuit below.

Explanation:

3 0
3 years ago
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