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antoniya [11.8K]

1 year 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]1 year ago

5 0

The answer is 7 because I just took the test!

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A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is

jonny [76]

Answer:

a) $t = 277.477\,s\,(4.625\min)$ , b) $v_{f} = 0\,\frac{mi}{h}$ , c) $a = -0.128\,\frac{ft}{s^{2}}$

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

$a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}$

$a = -0.185\,\frac{ft}{s^{2}}$

The time required to travel is:

$t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }$

$t = 277.477\,s\,(4.625\min)$

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be $v_{f} = 0\,\frac{mi}{h}$ .

c) The final constant deceleration is:

$a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}$

$a = -0.128\,\frac{ft}{s^{2}}$

7 0

3 years ago

The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg

kiruha [24]

Answer:

percentage change in volume = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × $10^{11}$ N/m²

bulk stress = $10^{7}$ N/m²

solution

we will apply here bulk modulus formula that is

bulk modulus = $\frac{bulk\ stress}{bulk\ strain}$ ...............1

put here value and we get

3.5 × $10^{11}$ = $\frac{10^7}{bulk\ strain}$

solve it we get

bulk strain = 2.8571 × $10^{-5}$

and

bulk strain = $\frac{change\ volume}{original\ volume}$

so that percentage change in volume is = 2.8571 × $10^{-5}$ × 100

percentage change in volume = 0.00285 %

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3 years ago

Pedro holds a heavy science book over his head for 10 minutes. Petro is doing work during that time. True or False

algol [13]

Answer:

True because he is working his arms to lift and hold the weight

Explanation:

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2 years ago

Vital role of maritime english among seaferers

seropon [69]

Answer:

uehgeg7djw7heidiisosowiuisiejei2k

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3 years ago

3. Which of these instruments is used to measure wind speed? A. anemometer C. wind sock B. thermometer D. wind vane It is an ins

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Answer:

wind vane if it can be used to show wind speed and the other is a

Explanation:

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2 years ago