All categories

3 years ago

The two strength factors that relate to all three competitive forces are

Physics

1 answer:

notka56 [123]3 years ago

7 0

I believe your answer is: Switching Costs & Customer Loyalty

You might be interested in

Which copper wire would have the lowest resistance?

Ronch [10]

B. The resistance is directly proportional to length and inversely proportional to cross sectional area

6 0

2 years ago

Read 2 more answers

How long( in hours, will it take for 500 000 C of charge to flow through a diode if it requires

Alina [70]

Answer:

277.78 hours

Explanation:

The formula for calculating the amount of charge is expressed as;

Q = It

I is the current

t is the time

Given

I =0.05A

Q = 50,000C

Required

Time t

Recall that: Q = It

t = Q/I

t = 50,000/0.05

t = 1,000,000secs

Convert to hours

1,000,000secs = 1,000,000/3600

1,000,000secs = 277.78 hours

Hence it will take 277.78 hours for the charge to flow through the diode

6 0

2 years ago

A Carnot engine operates between reservoirs at 600 and 300 K. If the engine absorbs 100 J per cycle at the hot reservoir, what i

PIT_PIT [208]

Answer:

W =50 J

Explanation:

given data:

T_h=600 k

T_L=300 k

Q_h=100 J

required:

<u>solution:</u>

║W║=║Q_h║(1-T_L/T_h)

=50 J

7 0

2 years ago

Read 2 more answers

A surveyor is using a magnetic compass 5.6 m below a power line in which there is a steady current of 140 A. (a) What is the mag

ArbitrLikvidat [17]

Answer:

(a) B = 5.6 micro Tesla

Explanation:

Current in the wire, i = 140 A

distance, r = 5 m

The formula for the magnetic field at a distance r due to the current carrying wire

$B=\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}$

$B=10^{-7}\times \frac{2\times140}{5}$

B = 5.6 x 10^-6 Tesla

B = 5.6 micro Tesla

(b) As the magnetic field of earth at this site is 20 micro tesla so the magnetic field due to current carrying wire interfere the magnetic compass.

4 0

2 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

4 0

2 years ago