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elena-s [515]
3 years ago
9

Explain in details the concept of economic geography​

Physics
1 answer:
vova2212 [387]3 years ago
3 0

Answer:

In the words of Hartshorn and Alexander: “Economic Geography is the study of the spatial variation on the earth’s surface of activities related to producing, exchanging and consuming goods and services. Whenever possible the goal is to develop generalizations and theories to account for these spatial variations.”

Explanation:

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According to Coulomb's law, if the separation between two particles of the same charge increases four times, the potential energ
Lera25 [3.4K]

Answer:

c.   V = k Q1 * Q2 / R1      potential energy of Q1 and Q2 separated by R

V2 / V1 = (R1 / R2) = 1/4

V2 = V1 / 4

8 0
3 years ago
An object moving 20 m/s
yan [13]

Answer:

<u>We are given:</u>

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

<u />

<u>Solving for Displacement:</u>

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

7 0
3 years ago
A bowling ball rolled with a force of 20 N accelerates at a rate of 5 m/sec2, a second ball rolled with the same force accelerat
emmasim [6.3K]

Answer:

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Explanation:

IM SO SORRY

5 0
3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S <br>​
Tems11 [23]

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}

6 0
3 years ago
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