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2 years ago

Can anyone explain how a Halbek Device works

Engineering

2 answers:

dedylja [7]2 years ago

8 0

Answer:

The Halbek Device can be used effectively on some weapons with practice and certain loadouts. ... This tends to help high-damage weapons or weapons with high multipliers

VARVARA [1.3K]2 years ago

4 0

Answer:

In Phantom Forces, the Halbek Device is a one-of-a-kind attachment. As stated in the description, the attachment effectively reverses, or rather inverses, the recoil values, causing the weapon to kick downwards instead of upwards.

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A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft

tatuchka [14]

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ; f_{Lw}=6.6 mi/h

5' on right ; f_{Lc} = 0.4 mi/hr

3' on left ; no adjustment

3 lanes in each direction f n = 3 mi/h

$v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}$

$f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$

$f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}$

= 0.877

$v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}$

= 1,555 veh/hr/lane

$FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}$

= (55 + 5) - 6.6 - 0.4 -3 -0

= 50 mi/h

$D = \dfrac{V_P}{s}$

$D = \dfrac{1555}{55} =28.27$

level of service is D using speed flow curves and LOS for basic free moving of vehicle

5 0

3 years ago

A 1-mm-diameter methanol droplet takes 1 min for complete evaporation at atmospheric condition. What will be the time taken for

Svet_ta [14]

Answer:

Time taken by the $1\mu m$ diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = $1\times 10^{- 6} m$

Diameter of the smaller droplet, R' = $0.5\times 10^{- 6} m$

Now,

Volume of the droplet, V = $\frac{4}{3}\pi R^{3}$

Volume of the smaller droplet, V' = $\frac{4}{3}\pi R'^{3}$

Volume of the droplet ∝ Time taken for complete evaporation

Thus

$\frac{V}{V'} = \frac{t}{t'}$

where

t' = taken taken by smaller droplet

$\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}$

$\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}$

t' = $60\times 10^{- 9} s = 60 ns$

5 0

3 years ago

Q1: The first option should always be to get out safely (RUN)

nekit [7.7K]

Answer:

Q1 true

Q2 true

And other I am confuse

6 0

2 years ago

An amplifier with 40 dB of small-signal, open-circuit voltage gain, an input resistance of 1 MO, and an output resistance of 100

Vanyuwa [196]

convert 40db to standard gain

AL=10^40/20=100

calculate total voltage gain

=AL×RL/RL+Ri

=83.33

38.41 DB

calculate power

Pi=Vi^2/Ri Po=Vo^2/RL

power gain= Po/Pi

=13.90×10^6

3 0

3 years ago

Which of the following describes design components that deal with the outward appearance or beauty of an object?

lisov135 [29]

Aesthetic elements are the components that are added to the design to be considered pleasing to the eye.

<h3>What are aesthetic elements?</h3>

They are those characteristics of an object that deal with the outward appearance or beauty of an object, that is, they are those elements that make it valuable, appreciable, relevant or transcendent.

To do this, the qualities must be in the design of the object but must also be perceived by the consumer, the aesthetic being what we like to perceive in objects.

Therefore, we can conclude that aesthetic elements are the components that are added to the design to be considered pleasing to the eye.

Learn more about aesthetic elements here: brainly.com/question/24568271

7 0

2 years ago

Read 2 more answers