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2 years ago

Which tool is used for cutting bricks and other masonry materials with precision?

Engineering

1 answer:

Llana [10]2 years ago

4 0

Answer:

A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.

Explanation:

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A cylindrical metal specimen having an original diameter of 11.34 mm and gauge length of 53.3 mm is pulled in tension until frac

WINSTONCH [101]

Answer:

a) 70.29 %

b) 37%

Explanation:

percent reduction can be found from:

PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2

= 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2

=70.29 %

percent elongation can be found from:

EL =L_f - Lo/Lo*100

= (73.17 -53.3/53.3)*100

= 37%

5 0

3 years ago

When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th

Amanda [17]

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

$\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant$

Differentiating the above equation:

$\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\$

$\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\$

$= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}$

3 0

2 years ago

The complex power of a load is 10-10j VA. What component should be added in parallel with the load so that the new load has a un

Korolek [52]

Need points bdjdjdhdhd

7 0

3 years ago

What is relation of crankshaft and camshaft

Monica [59]

Mierda me dices solo necesito puntos

6 0

2 years ago

Read 2 more answers

A 15,000 kg satellite is orbiting at 800 km above the Earth’s surface. What is the potential energy of the satellite?

vredina [299]

Answer:

$-8.34\times 10^{11}\ \text{J}$

Explanation:

m = Mass of satellite = 15000 kg

h = Distance above Earth = 800 km

R = Radius of Earth = 6371 km

M = Mass of Earth = $5.972\times 10^{24}\ \text{kg}$

G = Gravitational constant = $6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2$

Potential energy is given by

$U=-\dfrac{GMm}{R+h}\\\Rightarrow U=-\dfrac{6.674\times 10^{-11}\times 5.972\times 10^{24}\times 15000}{(6371+800)\times 10^3}\\\Rightarrow U=-8.34\times 10^{11}\ \text{J}$

The potential energy of the satellite is $-8.34\times 10^{11}\ \text{J}$ .

6 0

2 years ago