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1 year ago

Which of the given strategies is specifically a competitive advantage sustainment strategy?

1 answer:

6 0

The strategy that is specifically a competitive advantage-sustaining strategy is differentiation. The correct option is B.

<h3>What is a sustainment strategy?</h3>

This success is provided through a sustainment approach, which encourages salespeople to extend their efforts. There are four business level-strategies.

Each team member builds a repeatable performance that results in incremental gains as they learn to maintain their abilities. An organizational advantage is creating a work environment around this idea.

Therefore, the correct option is B. differentiation.

To learn more about sustainment strategy, refer to the below link:

brainly.com/question/11887945

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Pls help!!! will give brainly!!!

gogolik [260]

Short-term goals are planned for the near future while long-term goals will take longer to complete. An example of a short-term goal is trying to get into a certain college. A long-term goal is wanting to have a big family.

3 0

3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago

Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressur

Veseljchak [2.6K]

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

Initial temperature T1 = 27°C

Initial pressure P1 = 100 kPa

final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

Total work done W is determined where there is volume change which from point 2 to 3.

W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

7 0

4 years ago

Supercharging is the process of (a) Supplying the intake of an engine with air at a density greater than the density of the surr

iVinArrow [24]

Answer:

a)supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere

Explanation:

Supercharging is the process of supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere.

By doing this , it increases the power out put and increases the brake thermal efficiency of the engine.It also increases the volumetric efficiency of the engine.

So the our option a is right.

4 0

3 years ago

How a single force is resolved along the perpendicular axis acting an angle theta with horizontal?

Nataly [62]

Answer:

A single force, which is acting at angle θ from a horizontal axis, can be resolved into components which act along the perpendicular axis.

Consider the perpendicular axis x and y, where x represents the horizontal axis and y represents vertical axis.

The Force is resolved into 2 parts, one acts along x-axis and is represent by X. The other acts along y-axis and is represented by Y.

From the diagram we can see that the Force and its components X and Y makes up a right angles triangle, where θ is the angle from the x-axis

We know that:

cosθ = Base/Hypotenuse

cosθ = X/F

X = Fcosθ

We know that:

sinθ = Perpendicular/Hypotenuse

sinθ = Y/F

Y = Fsinθ

<h3>Relation of Force and its Components:</h3>

Force F can be represent by:

F = Fcosθ (along x-axis) + Fsinθ (along y-axis)

As they form a right angled triangle, we can use Pythagoras Theorem to show the relation between Force and its components.

Hypotenuse² = Base² + Perpendicular²

F² = X² + Y²

F² = (Fcosθ)² + (Fsinθ)²

$F = \sqrt{(F\cos\theta)^2+(Fsin\theta)^2}$

Where θ can be found by using any of the trignometric functions.

6 0

3 years ago