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ziro4ka [17]
3 years ago
11

__________ theory argues that city growth is generated by the pressure from the city center to expand outward. Expansion threate

ns to encroach on the surrounding areas and does so in concentric waves, or circles, with the center being the most intense, having the highest density and highest occupancy. These concentrations become progressively less intense and of lower density with greater distance from the center.
Physics
2 answers:
artcher [175]3 years ago
8 0

Answer: Concentric zone

Explanation:

PIT_PIT [208]3 years ago
5 0

Answer:

Defensible space

Explanation:

The defensible space theory states that an area can only be defined as a housing development if the people living in that location decide and defend that that area should adopt a good design and structure. For this reason, the area becomes a defensible space. In other words, the theory states that a residential area is only a successful defensible space, if the area forms a set in agreement between society and the physical elements of the region.

In addition, the theory argues that citizens should feel safe and responsible for this region, living in society and harmony.

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If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude
AveGali [126]

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

7 0
3 years ago
A ball is thrown upward with a speed of 40 m/s. Approximately how much time goes it take the ball to travel from the release loc
yulyashka [42]

Answer:

(I). The time at highest point 4.0 sec.

(II).  It returns to back to its original height in 8.1 sec

Explanation:

Given that,

Velocity v=40\ m/s

(I). We need to calculate the time at highest point

Using equation of motion

v=u+gt

Where, v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

Put the value into the formula

0=40-9.8t

-40=-9.8t

t=\dfrac{40}{9.8}

t=4.0\ sec

(II). We know that, when the ball to travel from the initial point and reached at initial point then the displacement is zero.

We need to calculate the total time when it returns to back to its original height

Using equation of motion

s=ut-\dfrac{1}{2}gt^2

Where, s = displacement

g = acceleration due to gravity

t = time

u = velocity

Put the value in the equation

0=40t-\dfrac{1}{2}9.8t^2

80t-9.8t^2=0

t=0, 8.1 sec

Hence. (I). The time at highest point 4.0 sec.

(II). It returns to back to its original height in 8.1 sec

5 0
3 years ago
A 62kg box is lifted 12 meters off the ground. How much work is done?
Temka [501]

Answer: 7291.2 joules

Explanation:

Work is done when force is applied on an object over a distance.

Thus, Workdone = Force X distance

Since Distance moved by box = 12 metres

mass of box = 62kg

Acceleration due to gravity when box was lifted is represented by g = 9.8m/s^2

Recall that Force = Mass x acceleration due to gravity

i.e Force = 62kg x 9.8m/s^2

= 607.6 Newton

So, Workdone = Force X Distance

Workdone = 607.6 Newton X 12 metres

Workdone = 7291.2 joules

Thus, 7291.2 joules of work was done.

4 0
3 years ago
5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj
irina [24]

Answer:

jack JACKKK

Explanation:

OK YAHHHHH SO DO THIS

3 0
4 years ago
A still ball of mass 0.514kg is fastened to a cord 68.7cm long and is released when the cord is horizontal. At the bottom of its
Alex787 [66]

Answer:

1.21 m/s

Explanation:

From the law of conservation of energy,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

where U₁ = initial potential energy of system =initial potential energy of still ball = mgh where m = mass of still ball = 0.514 kg, g = acceleration due to gravity = 9.8 m/s² and h = height = length of cord = 68.7 cm = 0.687 m.

K₁ = initial kinetic energy of system = 0

E₁ = initial internal energy of system = unknown and

U₂ = final potential energy of system = 0

K₁ = final kinetic energy of system = final kinetic energy of ball + steel block = 1/2(m + M)v² where m = mass of still ball, M = mass of steel block = 2.63 kg and v = speed of still ball + steel block

E₁ = final internal energy of system = unknown

So,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂

mgh = 1/2(m + M)v² + (E₂ - E₁)

Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4

Thus, mgh = 1/2(m + M)v² + (E₂ - E₁)

mgh = 1/2(m + M)v² + (m + M)v²/4

mgh = 3(m + M)v²/4

So, making v subject of the formula, we have

v² = 4mgh/3(m + M)

taking square root of both sides, we have

v = √[4mgh/3(m + M)]

Substituting the values of the variables into the equation, we have

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/{3(0.514 kg + 2.63 kg)}]

v = √[13.8422/{3(3.144 kg)}]

v = √[13.8422 kgm/s²/{9.432 kg)}]

v = √(1.4676 m²/s²)

v = 1.21 m/s

6 0
3 years ago
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