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3 years ago

You guys are amazing :D

Engineering

2 answers:

iren [92.7K]3 years ago

4 0

Thank you. You’re amazing as well !

Sloan [31]3 years ago

3 0

Ik i am thank you tho xoxo

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A 5000-lb truck is being used to lift a 1000-lb boulder B that is on a 200-lb pallet A. Knowing that the truck starts from rest

artcher [175]

Answer:

Explanation:

Total weight being moved = 5000+1000+200

= 6200 lb .

Force applied = 700 lb

= 700 x 32 = 22400 poundal .

acceleration (a) = 22400 / 6200

= 3.613 ft /s²

To know velocity after 6 ft we apply the formula

v² = u² + 2as

v² = 0 + 2 x 3.613 x 6

43.356

v = 6.58 ft/s

4 0

3 years ago

A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round

Harlamova29_29 [7]

Answer:

$\frac{Q_{square}}{Q_{circle}} = 0.785$

Explanation:

given data

types of drinking straws

square cross-sectional shape
round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square = perimeter of circle

4 × S = π × D

here S is length and D is diameter

S = $\frac{\pi D}{4}$ ....................1

and

ratio of flow rate through the square and circle is here

$\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}$

$\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}$

$\frac{Q_{square}}{Q_{circle}} = 0.785$

4 0

3 years ago

A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate

kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0

3 years ago

Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w

ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0

3 years ago

Can you guys please introduce yourself

EleoNora [17]

Answer: why?

Explanation:

6 0

3 years ago

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