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3 years ago

What is the example of indirect competition for resources

2 answers:

8 0

McDonald's and Subway would be an example of indirect competition because they do not sell the same product so they don't compete with each other compared to McDonald's and Wendy's.
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OverLord2011 [107]3 years ago

3 0

When something/someone is competing indirectly for something.

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A rigid tank contains water at 200kPa and an unknown temperature. When the tank is 100 °C the vapour starts condensing. Calculat

ipn [44]

Answer:

T = 450 C

Explanation:

given,

pressure of rigid tank containing water = 200 kPa

water start condensing at 100 °C

vapor saturation pressure at 100 °C = 101421 Pa (from saturation table)

specific volume = 1.6718 m³/kg

From super heated steam table

steam table is two set of table which is of energy transfer properties of water and steam.

at v = 1.6718 m³/kg and pressure = 200 kPa

temperature is equal to T = 450 C

7 0

3 years ago

A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

3 0

4 years ago

Help needed for science homework

Gemiola [76]

The answer is 1/2 * base * height

8 0

3 years ago

A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %

olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground = 46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt² ..............1

here d is distance and g is acceleration and t is time

so when object falling it will be

h = 4.9 t² ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)² ...................3

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²

t = 0.576 s and 3.771 s

we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h = 69.68 m

so height is 69.68 m

6 0

4 years ago

When sound is created it travels to the car through a

Rasek [7]

A sound wave is a longitudinal wave caused by vibrations and carried through a substance. The particles of the substance, such as air particles, vibrate back and forth along the path that the sound waves travel. Sound is transmitted through the vibrations and collisions of the particles.

This could maybe help you with your answer.

4 0

3 years ago