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meriva
3 years ago
9

What have you learned from the previous lesson? Let's try to check your prior knowledge

Engineering
1 answer:
Alla [95]3 years ago
3 0

........................................................................

Explanation:

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One horsepower is the ability to lift 550 pounds a height of one foot in one second or one horsepower is equal to 550ft-lb/s. Ho
madam [21]

Answer:

32 horsepower will be equal to 17600 ft-lb/sec

Explanation:

We have given 1 horsepower = 550 ft-lb/sec

We have to convert 32 horsepower into ft-lb/sec

As we know that 1 horsepower is equal to 550 ft-lb/sec

So for converting horsepower to ft-lb/sec we have multiply with 550

We have given 32 horse power

So 32\ horsepower =32\times 550=17600ft-lb/sec

So 32 horsepower will be equal to 17600 ft-lb/sec

5 0
4 years ago
4.A compression spring has a diameter of 0.67 in and a coil diameter of 0.067 in, with 30 total coils. The spring is to be made
Akimi4 [234]

Answer:

Fatigue factor of safety is 2.0267

Explanation:

Solution is attached below.

4 0
4 years ago
The flow of a real fluid has (more —less - same ) complexity of an ideal fluid, owing to the phenomena caused by the existence o
Viefleur [7K]

Answer:

The flow of a real fluid has <u>more</u> complexity as compared to an ideal fluid owing to the phenomena caused by existence of <u>viscosity</u>

Explanation:

For a ideal fluid we know that there is no viscosity of the fluid hence the boundary condition need's not to be satisfied and the flow occur's without any head loss due to viscous nature of the fluid. The friction of the pipe has no effect on the flow of an ideal fluid. But for a real fluid the viscosity of the fluid has a non zero value, the viscosity causes boundary layer effects, causes head loss and also frictional losses due to pipe friction hugely make the analysis of the flow complex. The losses in the energy of the flow becomes complex to calculate as frictional losses depend on the roughness of the pipe and Reynolds number of the flow thus increasing the complexity of the analysis of flow.

3 0
3 years ago
1.The moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kN/m^3, degree of saturati
alexandr1967 [171]

Answer:

Gs = 2.647

e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}

where,  \gamma_m = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

\gamma_w = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation,\gamma_m = 16.62 kN/m3

puttng all value to get Gs value;

16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}

Gs - 1.055*e = 1.805 .........(2)

solving both  equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

6 0
3 years ago
Which of the following are not related to a materials structure? (Mark all that apply) a)- Atomic bonding b)- Crystal structure
belka [17]

The answer is c) atomic number

3 0
3 years ago
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