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2 years ago

Determine how much concrete you will need for a slab which is 50 feet by 30 feet wide and 1 foot thick

Engineering

1 answer:

Levart [38]2 years ago

6 0

Answer:

Amount of concrete need to make slab = 1,500 feet³

Explanation:

Given:

Length of slab = 50 feet

Width of slab = 30 feet

Height of slab = 1 feet

Find:

Amount of concrete need to make slab

Computation;

Amount of concrete need to make slab = Volume of cuboid

Volume of cuboid = (l)(b)(h)

Amount of concrete need to make slab = (50)(30)(1)

Amount of concrete need to make slab = 1,500 feet³

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2 years ago

Express the following quantities to the nearest standard prefix using no more than three digits.(a) 20,000,000 Hz(b) 1025 W(c) 0

bija089 [108]

Answer:

(a) 20 MHz

(b) 1.025 KW

(d) 33 pF

Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = 20 MHz

(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = 1.025 KW

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3 years ago

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

mojhsa [17]

Answer:

A) Upper bound modulus of elasticity; E = 165.6 GPa

B) Lower bound modulus of elasticity; E = 83.09 GPa

Explanation:

A) Formula for upper bound modulus is given as;

E = E_m(1 - V_f) + E_f•V_f

We are given;

E_m = 60 GPa

E_f = 380 GPa

V_f = 33% = 0.33

Thus,

E = 60(1 - 0.33) + 380(0.33)

E = (60 x 0.67) + 125.4

E = 165.6 GPa

B) Formula for lower bound modulus is given as;

E = 1/[(V_f/E_f) + ((1 – V_f)/E_m)]

E = 1/[(0.33/380) + ((1 – 0.33)/60)]

E = 1/(0.0008684 + 0.01116667)

E = 1/0.01203507

E = 83.09 GPa

3 0

2 years ago

Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m^

Nat2105 [25]

Answer:

The answer is " $\bold{ 259.2 \times 10^{11} }$ ".

Explanation:

The amount of kilograms, which travel in a thick sheet of hydrogen:

$M= -DAt \frac{\Delta C}{ \Delta x} \\\\$

$D =1.0 \times 10^{8} \ \ \ \frac{m^2}{s} \\\\ A = 0.20 \ m^2\\\\t = 1\ \ h = 3600 \ \ sec \\\\$

calculating the value of $\Delta C:$

$\Delta C =C_A -C_B$

$= 2.4 - 0.6 \\\\ = 1.8 \ \ \frac{kg}{m^3}$

calculating the value of $\Delta X:$

$\Delta x = x_{A} -x_{B}$

$= 0 - (5\ mm) \\\\ = - 5 \ \ mm\\\\= - 5 \times 10^{-3} \ m$

$M = -(1.0 \times 10^{8} \times 0.20 \times 3600 \times (\frac{1.8}{-5 \times 10^{-3}})) \\\\$

$= -(1.0 \times 10^{8} \times 720 \times (\frac{1.8}{-5 \times 10^{-3}})) \\\\= -(1.0 \times 10^{8} \times \frac{ 1296}{-5 \times 10^{-3}})) \\\\= (1.0 \times 10^{8} \times 259.2 \times 10^3)) \\\\= 259.2 \times 10^{11} \\\\$

3 0

3 years ago