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3 years ago

6

. In order to prevent injury from inflating air bags, it is recommended that vehicle occupants sit at least __________ inches aw

ay from the air bag.

2 answers:

scZoUnD [109]3 years ago

8 0

10 inches is the answer

Umnica [9.8K]3 years ago

8 0

10 inches is the answer to that question sir

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It is not a practical proposition to take direct measurements in nanoscale, but we can estimate variations in position and momen

Volgvan

Answer:

Answer is c Heisenberg's uncertainty principle

Explanation:

According to Heisenberg's uncertainty principle there is always an inherent uncertainty in measuring the position and momentum of a particle simultaneously.

Mathematically

$\Delta x\times \Delta \overrightarrow{p}\geq \frac{h}{4\pi }$

here 'h' is planck's constant

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3 years ago

An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8

marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

P= 800 x 108.3

P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

$a=18.66\ m/s^2$

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0

4 years ago

As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de

katen-ka-za [31]

yes it will

Explanation:

5 0

4 years ago

A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,

disa [49]

Answer: (a) 36.18mm

(b) 23.52

Explanation: see attachment

4 0

4 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

3 0

3 years ago