What is the primary function of the small intestines

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area $A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2$

Let the tensile modulus of Nickel $E = 170 \times 10^9Pa$ .

The elongation of the rod can be calculated using the following formula:

$\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m$

6 0

3 years ago

you know that there are 1609 meter in a mile. the number of feet in a mile is 5280. what is the speed snail from problem 7 per m

Zanzabum

Answer:

We know that 1 meter = 100 centimeters, and 1 foot = 12 inches.

So (1,609 meters) x (100 centimeters/meter) = (5,280 feet) x (12 inches/foot)

The second fraction on each side of the equation is equal to ' 1 ', because

the numerator is equal to the denominator, so sticking it in there doesn't

change the value of that side of the equation. But now we can cancel some

units,and wind up with the units we need.

(1,609 meters) x (100 centimeters/meter) = (5,280 feet) x (12 inches/foot)

(1,609 x100) centimeters = (5,280 x 12) inches

160,900 centimeters = 63,360 inches

Divide each side by 63,360 : 2.54 centimeters = 1 inch

Explanation:

5 0

2 years ago

Why is a minimum of three seismic stations needed to find the epicenter of an earthquake?

vazorg [7]

Each station can detect how far away the epicenter was. So each station basically has a circle made of possible epicenters. When you have three, you narrow it down to one, final point.

5 0

3 years ago

One reason why display rules differ across cultures is because ___________. A. values and beliefs differ across cultures B. some

Soloha48 [4]

Answer:

The correct answer would be C, certain display rules are more appropriate than others.

Explanation:

I hope this helps you:)

6 0

3 years ago

An isloated point charge produce an electric field with magnitude E at a point 2 m away. At a point 1 m from the charge magnitud

Nina [5.8K]

Answer:

the correct answer is C, E’= 4E

Explanation:

In this exercise you are asked to calculate the electric field at a given point

E = $k \frac{q}{r^2}$

indicates that the field is E for r = 2m

E = $\frac{ k q}{4}$ (1)

the field is requested for a distance r = 1 m

E ’= k \frac{q}{r'^2}

E ’= k q / 1

from equation 1

4E = k q

we substitute

E’= 4E

so the correct answer is C

8 0

3 years ago