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3 years ago

The volume of the pyramid is 36 cubic cm, find the volume of the prism.

Engineering

1 answer:

ser-zykov [4K]3 years ago

5 0

Answer:

Given :- the volume of the pyramid is 36 cubic cm , find the volume of the prism on same base and same height as pyramid .

Answer :-

we know that,

Volume of pyramid = (1/3) * Base area * height .

Volume of prism = Base area * height .

so,

→ Volume of pyramid = 36 cm³

→ (1/3) * Base area * height = 36

→ Base area * height = 36 * 3

→ Base area * height = 108 cm³.

then,

→ Volume of prism = Base area * height .

→ Volume of prism = 108 cm³ (Ans.)

Explanation:

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Please read and an 3. Many Jacks use hydraullc power. A) O True B) O False

drek231 [11]

Answer:

A) True. I hope this helps

5 0

3 years ago

A large part in a turbine-generator unit operates near room temperature and is made of ASTM A470-8 steel ( ). A surface crack ha

mr Goodwill [35]

Answer: safety factor= 1.26

safety factor does not exceed 2. Therefore, it is not safe to operate.

Explanation:

Express the stress infinity factor (k) that relate to the crack length (Q) applied stress for the centre crack of the plate.

K= Fsq √ πa/Q

Here,

the dimension less function quality is F

Crack length is a

Q= 1 + 1.48(a/c)^1.65

The surface length is c

c= 25mm

a= 15mm

Substitute c= 25mm and a=15mm into the equation

Q= 1 + 1.48(15mm/25mm)^1.65

Q= 1 + 1.48(0.6)^1.65

Q= 1 + 0.630

Q= 1.630

Find the stress intensity factor of the corresponding ratio of 0.4 and 0.12

Where

F= 1.12

S= 250mpa

Q= 1.63

a= 15mm

Substitute into the qequation

K= (1.12) (250mpa) √π(15mm× (1/1000mm))/1.63

K= (1.12) (250mpa)√47.1mm× (1/1000mm))/1.63

K= 2.80mpa √0.0471× (1m)/1.63

K= 280mpa × 0.170√m

K= 476mpa√m

Calculate the safety

Xk= Kk/K

Where fraction toughness is Kk

From the table of fraction toughness, corresponding tensile property for metal at room temperature, select this fraction toughness for ASTM A470-8 steel.

Substitute 47.6mpa√m for k

60mpa√m for Kk

Xk= 60mpa√m/47.6mpa√m

Xk= 1.26

In conclusion, from the above result, safety factor does not exceed 2. Therefore, it is not safe to operate.

3 0

3 years ago

5) A 80-kg man has a total foot imprint area of 480 cm2. Determine the pressure this man exerts on the ground if (a) he stands o

attashe74 [19]

Answer:

The pressure exerted by this man on ground

(a) if he stands on both feet is 8.17 KPa

(b) if he stands on one foot is 16.33 KPa

Explanation:

(a)

When the man stand on both feet, the weight of his body is uniformly distributed around the foot imprint of both feet. Thus, total area in this case will be:

Area = A = 2 x 480 cm²

A = 960 cm²

A = 0.096 m²

The force exerted by man on his area will be equal to his weight.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.096 m²

P = 8166.67 Pa = 8.17 KPa

(b)

When the man stand on one foot, the weight of his body is uniformly distributed around the foot imprint of that foot only. Thus, total area in this case will be:

Area = A = 480 cm²

A = 0.048 m²

The force exerted by man on his area will be equal to his weight, in this case, as well.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.048 m²

P = 16333.33 Pa = 16.33 KPa

4 0

3 years ago

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after

kotegsom [21]

This question is incomplete, the complete question is;

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after 100 s, the reaction is 40% complete, how long (total time in seconds) will it take the transformation to go to 95% completion

y = 1 - exp( -ktⁿ )

Answer: the time required for 95% transformation is 242.17 s

Explanation:

First, we calculate the value of k which is the dependent variable in Avrami equation

y = 1 - exp( -ktⁿ )

exp( -ktⁿ ) = 1 - y

-ktⁿ = In( 1 - y )

k = - In( 1 - y ) / tⁿ

now given that; n = 2, y = 40% = 0.40, and t = 100 s

we substitute

k = - In( 1 - 0.40 ) / 100²

k = - In(0.60) / 10000

k = 0.5108 / 10000

k = 0.00005108 ≈ 5.108 × 10⁻⁵

Now calculate the time required for 95% transformation

tⁿ = - In( 1 - y ) / k

t = [- In( 1 - y ) / k ]^1/n

n = 2, y = 95% = 0.95 and k = 5.108 × 10⁻⁵

we substitute our values

t = [- In( 1 - 0.95 ) / 5.108 × 10⁻⁵ ]^1/2

t = [2.9957 / 5.108 × 10⁻⁵]^1/2

t = [ 58647.22 ]^1/2

t = 242.17 s

Therefore the time required for 95% transformation is 242.17 s

8 0

3 years ago

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$