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3 years ago

Sarah is on a see saw with her friend. The pivot or fixed point where the bar is free to move is called the _____________?

Engineering

1 answer:

Triss [41]3 years ago

3 0

Answer: Fulcrum

Definition:
the point on which a lever rests or is supported and on which it pivots.

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3 years ago

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Explanation:

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2 years ago

Given a books table with columns named book_title and year_released, the following SQL statement will select the number of books

lord [1]

Question Continuation

True or False?

Answer:

False

Explanation:

Given

Columns: book_title and year_released

SQL statement: SELECT book_title, year_released FROM books GROUP BY year_released

Expected Result: To select the number of books released each year

Although the GROUP BY used in the SQL statement is meant to group rows with same values into summary rows but it's not user properly because the GROUP BY clause, as used here needs the COUNT aggregate functions to group the result-set by one or more columns.

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3 years ago

Explain ten application of solar electric energy

Aleksandr-060686 [28]

Answer:

Some of the major application of solar energy are as follows: (a) Solar water heating (b) Solar heating of buildings (c) Solar distillation (d) Solar pumping (e) Solar drying of agricultural and animal products (f) Solar furnaces (g) Solar cooking (h) Solar electric power generation (i) Solar thermal power production (j) Solar green houses.

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2 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

$y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\$

Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

3 0

4 years ago