All categories

3 years ago

A benefit to using the medium the author used in "Great Rock and Roll

Engineering

2 answers:

lana66690 [7]3 years ago

7 0

Incomplete question. The options read;

A. can change the story's ending

B. listens to the dialogue

C. hears the rock songs

D. feels more connected to the text.

Answer:

D. feels more connected to the text.

Explanation:

Remember, the underlying motive of an author when writing any text is to make his readers or audience more connected to the material been read or heard.

Hence, in "Great Rock and Roll Pauses" we can conclude that the author's choice of medium was motivated by the same desire of making the audience feel more connected to the text.

IgorLugansk [536]3 years ago

6 0

Answer:

The audience can actually hear the music.

Explanation: dont listen to below average iq people :0

You might be interested in

A 625 g basketball and a 58.5 g tennis ball are dropped from a height of d = 1.5 m onto the floor. The coefficient of restitutio

stich3 [128]

Answer:

Maximum height=7.3535 m

Explanation:

Solution of the problem is given in the attachments.

3 0

3 years ago

Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat

Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 = 0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = $\frac{m}{\rho}$

V = Av = $\frac{\pi}{4} * d^2 * v1$

V = $\frac{\pi}{4} * 0.28^2 * 5$

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = $\frac{V}{v}$ = $\frac{0.30787}{0.11418}$

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = $\frac{v2}{A2}$

v2 = $\frac{0.3705}{\frac{\pi}{4} * 0.28^2}$

v2 = 6.017 m/s

7 0

3 years ago

Casein, a dairy product used in making cheese, contains 25% moisture when wet. A dairy sells this product for $40/100 kg. If req

Nataly_w [17]

Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.

<h3>What is the mass of casein in wet casein?</h3>

The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg

Mass of water 250 kg

The mass of casein is constant while the moisture content can be changed.

At 12% moisture content;

750 kg = 88%%

100 % = 100 ×750/88 = 852.27 kg

Therefore, the mass of dried casein produced os 852.3 kg.

Learn more about mass at: brainly.com/question/24658038

#SPJ1

3 0

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

3 years ago

A NC drill press is to perform a series of through-hole drilling operations on a 1.75 in thick aluminum plate that is a componen

jekas [21]

Answer:

26.7 min

Explanation:

First, we will find the time required to drill each hole:

N = 300 x 12/0.75 $\pi$ = 1527.7 rev/min

fr = 1527.7 (0.015) = 22.916 in/min

Formula for distance per hole: 0.5 + A + 1.75

A = 0.5 (0.75) tan (90-100 / 2) = 0.315 in
Tm = (0.5 + 0.315 + 1.75) / 22.916 = 0.112 min

Now, we will calculate the time required to draw back the drill form hole:

= 0.112 / 2 = 0.056 min

Time to move between holes = 1.5 / 15 = 0.1 min

For 100 holes, the number of moves between holes = 99

Total time required to drill 100 holes (t):

t = 100 (0.112 + 0.056) + 99 (0.1) = 26.7 min

7 0

3 years ago

Read 2 more answers