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Archy [21]
2 years ago
5

Price controls on rents are frequently implemented by governments in an effort to protect renters from high housing prices. Diff

erentiate the type of price control used from other potential price controls, and then formulate reasons as to why governments should be careful when implementing these types of controls.
PLZ I NEED HELP!!
Physics
1 answer:
Pavlova-9 [17]2 years ago
7 0

The most common price control is setting a fixed rent price to protect the tenants. There are other alternatives to this price control. Also, price control should be carefully implemented.

Rent regulation is a term that refers to a set of laws that focus on ensuring that individuals have access to rented housing.

The rental regulation system has a principe that is:

  • Price controls: Set a rent price and set the increase in a percentage of it (the rent is $ 100 and the annual increase is 5%)

There are alternatives price control such as:

  • Price according to the type of property: Set a rental price according to the type of property and its benefits (for example pool, garage, garden, etc)
  • Price according to the number of tenants: Set a rental price according to the number of people who will inhabit the property

Generally, tenants are allowed to raise the value of the rent each year by a minimal proportion. This is necessary to avoid that the landlord does not increase the rental value in large amounts because the tenant will not have access to the rented house.

However, governments should be careful with this regulations because they can negative afect the landlord by limiting his rights over the property.

Learn more in:

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What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti
Nataliya [291]

Archimedes principle states that

 

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

 

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

 

F2 = (πr2^2 / πr1^2) * mg

 

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

 

Substituting the values to the derived equation, we get

 

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

 

 

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6 0
2 years ago
An aluminum wire is 5.87 m long and has a diameter of 6.07 mm.
antoniya [11.8K]

Answer: 5.72 x 10-3Ω

Explanation:

Hi, to answer this question, first we have to calculate the cross sectional area of the cable:

Diameter (D)=6.07 mm

Since: 1000mm = 1m

6.07mm/ 1000mm/m = 0.00607 meters

Area of a circle : π (d/2)^2

A = π (0.00607/2)^2= 0.000028937 m2

Resistance formula:

Resistance (R) = P(resistivity) L (length)÷A (cross sectional area )

Replacing with the values given:

R = (2.82x10-8 x 5.87) / 0.000028937

R = 5.72 x 10-3Ω

Feel free to ask for more if needed or if you did not understand something.

7 0
3 years ago
What's the frequency of a wave with a wavelength of 10 and velocity of 200m/s?
uranmaximum [27]

Answer:

\boxed {\boxed {\sf 20 \ Hz}}

Explanation:

The frequency of a wave can be found using the following formula.

f=\frac{v}{\lambda}

where <em>f</em> is the frequency, <em>v</em> is the velocity/wave speed, and λ is the wavelength.

The wavelength is 10 meters and the velocity is 200 meters per second.

  • 1 m/s can also be written as 1 m*s^-1

Therefore:

v= 200 \ m*s^{-1} \\\lambda = 10 \ m

Substitute the values into the formula.

f=\frac{200 \ m*s^{-1}}{10 \ m}

Divide and note that the meters (m) will cancel each other out.

f=\frac{200 \ s^{-1}}{10 \ }

f=20 \ s^{-1}

  • 1 s^-1 is equal to Hertz
  • Therefore, our answer of 20 s^-1 is equal to 20 Hz

f= 20 \ Hz

The frequency of the wave is <u>20 Hertz</u>

7 0
2 years ago
Two ping pong balls have been painted with metallic paint and charged by contact with an Van de Graaff generator. The charge on
andre [41]

Answer:

0.035 N

Explanation:

Parameters given:

Charge q1 = -3.31x10^(-7) C

Charge q2 = -5.7x10^(-7) C.

Distance between them, R = 22 cm = 0.22 m

Electrostatic force between to particles is given as:

F = (k* q1 * q2) / R²

F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²

F = 0.035 N

6 0
3 years ago
Read 2 more answers
Assuming that each of the following objects is a typical example of its class, rank them by increasing density.
inysia [295]

molecular cloud <interstellar cloud <1 Msun protostar <1 Msun star <intercloud gas

Explanation:

<u>Molecular cloud-</u> They are a variety of interstellar cloud in which molecular hydrogen can sustain themselves. They have a very low temperature ranging from -440 to -370 degrees Fahrenheit or between<u> 10 to 50 Kelvin. </u>Owing to their extremely low temperature, they appear mostly dark when viewed through telescopes.

<u>Interstellar cloud-</u> They are a congregation of a large number of interstellar gases, dust and plasma in any galaxy or universe. They have varying temperature depending on their proximity to a star. E.g. Neutral hydrogen atom clouds have a temperature of around <u>just 100 Kelvin</u> while those in the near vicinity of a star have temperatures as high as 10,000 Kelvin.

<u>1 Msun star-</u> These stars have temperature anywhere between <u>5300 and 6000 Kelvin</u>. The main source of such high surface temperature is nuclear fusion process where elemental hydrogen molecules are fused to form helium molecules.  

<u>1 Msun protostar-</u> protostar is rather a young star which is still in formation phase (i.e. gathering mass from the parent molecular cloud). They have temperature anywhere between <u>2000-3000</u> kelvin and are accompanied by dust usually.

<u>Intercloud gas- </u>These are the remainder gases that are spread throughout the interstellar space. This Intercloud gas is divided into warm intercloud medium and extremely hot coronal gas with temperatures comparing to Sun’s corona. Warm intercloud forms the dominant part of intercloud gas with a temperature around <u>8000 Kelvin</u>.

8 0
3 years ago
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