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2 years ago

12

When your complex reaction time is compromised by alcohol, an impaired person's ability to respond to emergency or unanticipated

situations is greatly______.

1 answer:

babunello [35]2 years ago

3 0

Answer:

decreased

Explanation:

when impaired you react slower then you would sober.

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A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

$T_{BE}-W=0$ hence

$T_{BE}=W=20*9.81=196.2 N$

Therefore, tension in the cable, $T_{BE}=196.2 N$

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

$196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0$

$24.525-39.24+0.2D_x=0$

$D_x=73.575 N$

Similarly,

$A_x-D_y=0$

$A_x=73.575 N$

Therefore, both reactions at A and D are 73.575 N

7 0

2 years ago

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

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siniylev [52]

Answer:

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Explanation:

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