The federal highway administration reports nearly

A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is appli

Scrat [10]

Answer:

$X_t=2.17391304*10^{-4}$

$X_r=2.89855072*10^{-4}$

$e_t=0.0026$

$e_r=0.0035$

Explanation:

From the question we are told that:

Dimension $12*12$

Thickness $l_t=5mm=5*10^-3$

Normal tensile force on top side $F_t= 15kN$

Normal tensile force on right side $F_r= 20kN$

Elastic modulus, $E=115Gpap=>115*10^9$

Generally the equation for Normal Strain X is mathematically given by

$X=\frac{Force}{Area*E}$

Therefore

For Top

$X_t=\frac{Force_t}{Area*E}$

Where

$Area=L*B*T$

$Area=12*10^{-2}*5*10^{-3}$

$Area=6*10^{-4}$

$X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}$

$X_t=2.17391304*10^{-4}$

For Right side $X_r=\frac{Force_r}{Area*E}$

Where

Area=L*B*T

$Area=12*10^{-2}*5*10^{-3}$

$Area=6*10^{-4}$

$X_r=2.89855072*10^{-4}$

Generally the equation for elongation is mathematically given by

$e=strain *12$

For top

$e_t=2.17391304*10^{-4}*12$

$e_t=0.0026$

For Right

$e_r=2.89855072*10^{-4} *12$

$e_r=0.0035$

5 0

3 years ago

A pipe leads from a storage tank on the roof of a building to the ground floor. The absolute pressure of the water in the storag

Fudgin [204]

Answer:

6.4 m/s

Explanation:

From the equation of continuity

A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively

$A1=\pi (r1)^{2}$

$A2=\pi (r2)^{2}$

where r1 and r2 are radius of inlet and outlet respectively

v1 is given as 1.6 m/s

Therefore

$\pi (0.01)^{2}\times 1.6 = \pi (0.005)^{2}v2$

$V2=\frac {\pi (0.01)^{2}\times 1.6}{\pi (0.005)^{2}}=6.4 m/s$

3 0

4 years ago

7. What happens to the current in a parallel circuit?|

Bess [88]

Answer:

Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source. ... If one of the parallel paths is broken, current will continue to flow in all the other paths.

8 0

4 years ago

1. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle.

sammy [17]

Answer:

A. Yes

B. Yes

Explanation:

We want to evaluate the validity of the given assertions.

1. The first statement is true

The sine rule stipulates that the ratio of a side and the sine of the angle facing the side is a constant for all sides of the triangle.

Hence, to use it, it’s either we have two sides and an angle and we are tasked with calculating the value of the non given side

Or

We have two angles and a side and we want to calculate the value of the side provided we have the angle facing this side in question.

For notation purposes;

We can express the it for a triangle having three sides a, b, c and angles A,B, C with each lower case letter being the side that faces its corresponding big letter angles

a/Sin A = b/Sin B = c/Sin C

2. The cosine rule looks like the Pythagoras’s theorem in notation but has a subtraction extension that multiplies two times the product of the other two sides and the cosine of the angle facing the side we want to calculate

So let’s say we want to calculate the side a in a triangle of sides a, b , c and we have the angle facing the side A

That would be;

a^2 = b^2 + c^2 -2bcCosA

So yes, the cosine rule can be used for the scenario above

3 0

3 years ago

1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. T

Elenna [48]

Answer:

Concentration = 10.33 kg/m³

Explanation:

We are given;

Mass of solids; 10,000 kg

Volume; V = 440,000 L = 440 m³

Rate at which water is pumped out = 40,000 liter/h

Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³

Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;

200m³/440m³ = 5/11

Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;

Thus,

Amount left; = 10000 x (5/11) = 4545 kg

The concentration would be calculated by:

Concentration = amount left/initial volume

Thus,

Concentration = 4545/440 = 10.3 kg/m³

8 0

3 years ago