Ask question

Ask question

All categories

antoniya [11.8K]

2 years ago

7

If extension done in a stork position, during backward bending in a standing position of a special test for lumbar spine, produc

es pain in the lumbar or sacral regions of the spine, it may indicate _____.

1 answer:

Klio2033 [76]2 years ago

5 0

The answer is 7 because I just took the test!

You might be interested in

A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of 15°C and convection heat transfer

gulaghasi [49]

Answer:

809.98°C

Explanation:

STEP ONE: The first step to take in order to solve this particular Question or problem is to find or determine the Biot value.

Biot value = (heat transfer coefficient × length) ÷ thermal conductivity.

Biot value = (220 × 0.1)÷ 110 = 0.2.

Biot value = 0.2.

STEP TWO: Determine the Fourier number. Since the Biot value is greater than 0.1. Tis can be done by making use of the formula below;

Fourier number = thermal diffusivity × time ÷ (length)^2.

Fourier number = (3 × 60 × 33.9 × 10^-6)/( 0.1)^2 = 0.6102.

STEP THREE: This is the last step for the question, here we will be calculating the temperature of the center plane of the brass plate after 3 minutes.

Thus, the temperature of the center plane of the brass plane after 3 minutes = (1.00705) (0.89199) (900- 15) + 15.

= > the temperature of the center plane of the brass plane after 3 minutes = 809.98°C.

5 0

3 years ago

How would you describe what would happen to methane if the primary bonds were to break?

erastova [34]

Answer:

All the bonds in methane (CH4CH4) are equivalent, and all have the same dissociation energy.

The product of the dissociation is methyl radical (CH3CH3). All the bonds in methyl radical are equivalent, and all have the same dissociation energy.

The product of that dissociation is methylene (CH2CH2). All the bonds in methylene are equivalent, and all have the same dissociation energy.

The product of that dissociation is methyne (CHCH) .

The C-H bonds in methane do not have the same dissociation energy as C-H bonds in methyl radical, which in turn do not have the same dissociation energy as the C-H bonds in methylene, which are again different from the C-H bond in methyne.

If (by some miracle) you were able to get all four bonds in methane to dissociate absolutely simultaneously, they would all show the same dissociation energy… but that energy, per bond broken, would be different than the energy required to break just one C-H bond in methane, because the products are different.

(In this case, it’s CH4→C+4HCH4→C+4H versus CH4→CH3+HCH4→CH3+H.)

To alter hydrocarbons you add enough energy to break a C-H bond. Why does only one bond break? What concentrates the energy on one C-H bond?

the weakest CH bond is the one that breaks. in plain alkanes it has to do with the molecular orbital interactions between neighboring carbon atoms. look at propane for example. the middle carbon has two C-C bonds, and each of those C-C bonds is strengthened by slight electron delocalization from the C-H bonds overlapping with the antibonding orbitals of the adjacent carbons.

since the C-H bonds on the middle carbon donate electron density to both of its neighbors, those two are weakest.

one of them will break preferentially.

which one actually breaks depends on the reaction conditions (kinetics). frankly it's whichever one ramdomly approaches a nucleophile first. when the nucleophile pulls of one of the H's, the other C-H bonds start to share (delocalize) the negative charge across the whole molecule. so while the middle C feels the majority of the negative charge character, the other two C's take on a fair amount as well...

by the way, alkanes don't really like to break and form anions like that.

a better example would be something like isopropyl iodide, where the C-I bond breaks and the I carries away the electron pair, forming a carbocation (also not particularly stable, but more so than the carbanion).

7 0

3 years ago

Problem 34.3 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E

Natasha2012 [34]

The image is missing, so i have attached it.

Answer:

A) P = 65.11 KN

B) Q = 30 KN

Explanation:

We are given;

The end reaction of the beam; F = 100KN

Coefficient of static friction between two steel surfaces;μ_ss = 0.3

Coefficient of static friction between steel and concrete;μ_sc = 0.6

So, F1 = μ_ss•F =0.3 x 100 = 30 KN

F2 = μ_ss•N_EF = 0.3N_EF

From the screen shot, we see that the angle is 12°

Sum of forces in the Y-direction gives;

F2•sin12 - N_EF•cos12 + 100 = 0

Rearranging gives;

N_EF•cos12 - F2•sin12 = 100

Let's put 0.3N_EF for F2 to give;

N_EF•cos12 - 0.3N_EF•sin12 = 100

Thus;

N_EF(0.9158) - 0.1247 = 100

N_EF(0.9781) = 100 + 0.1247

N_EF = 100.1247/0.9158

N_EF = 109.33 KN

Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN

Wedge will move if;

P = (F1 + F2cos12 + N_EFsin12)

Thus;

P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)

P ≥ 65.11 KN

B) For static equilibrium, Q = F1

Thus, Q = 30 KN

3 0

3 years ago

Think of the differences between circuit-switching and packet-switching paradigms in the Internet core design. Assume an Interne

dem82 [27]

Answer:

0.264 ; 0.079

Explanation:

Given that:

Sample size, n = 100

Probability of being active, p = 1% = 1/100 = 0.01

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that more than 1 user will be active

P(x > 1) = 1 - [p(x=0) + p(x = 1)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x > 1) = 1 - [0.366 + 0.370]

P(x > 1) = 0.264

2.)

Probability that more than 2 user will be active

P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185

P(x > 1) = 1 - [0.366 + 0.370 + 0.185]

P(x > 1) = 0.079

7 0

3 years ago

A residential heat pump has a coefficient of performance of 1.49 How much heating effect, in kJ/h, will result when 4 kW is supp

AfilCa [17]

Answer:

21.456 kJ/h

Explanation:

See the figure attached. In this case

$W_{cycle} = 4 kW$

$Q_{out} = \text{heating effect}$

Coefficient of performance in heat pump is defined by

$COP = \frac{Q_{out}}{W_{cycle}}$

$Q_{out} =COP*W_{cycle}$

$Q_{out} =1.49*4 \, W$

$Q_{out} = 5.96 \, W$

Now it is necessary to change units, remember that Watt (W) is defined as J/s

$Q_{out} = 5.96 \frac{J}{s} \frac{3600s}{1 h} \frac{1 kJ}{1000 J}$

$Q_{out} = 21.456 \frac{kJ}{h}$

3 0

3 years ago