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geniusboy [140]

3 years ago

10

One principle of environmental law and policy in the U.S. is to make polluters pay . True or false ?

1 answer:

crimeas [40]3 years ago

5 0

In a way it’s true because you can get a ticket for getting caught littering

You might be interested in

Consider a spherical tank full of water with radius 3 m (plus a spout on top 1m high). Set up an integral expression for the wor

sweet-ann [11.9K]

Answer:

The answer to the question is

4433.416 kJ

See explanation below

(3-y)²+r² = 3² or

6y-y² = r²

r =√(6y-y²)

The volume of a small section of height Δy = Δy ×(√(6y-y²))²×π

For water with density of 1000 kg/m³, the mass of the slice

= 1000×Δy ×(√(6y-y²))²×π and since force = mass × acceleration we have

1000×Δy ×(√(6y-y²))²×π ×g = 1000×Δy ×(√(6y-y²))²×π ×9.81

The work done to move a unit height of y+1 = Force × Distance

W = 1000×Δy ×(√(6y-y²))²×π ×9.81 × (y+1)

Integrating the entire section of the sphere that is 2×r high, or from 0 to 6 we get

W = $\int\limits^6_0 {1000*(6y-y^{2} ) *\pi *9.81 * (y+1)} \, dy$

= $9810*\pi \int\limits^6_0 {5y^{2} -y^{3} +6y \, dy$

$= 9810*\pi *[\frac{5y^{3} }{3} -\frac{y^{4} }{4} +3y^{2} ]^{6} _{0}$

=9810×π×144 =4433416 J

7 0

2 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

3 years ago

What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

Eduardwww [97]

Answer:

Final temperature, $T_f=21.85^{\circ}$

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, $T_i=41^{\circ}C$

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, $c=0.235\ J/g\ C$

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

$Q=mc\Delta T$

$Q=mc(T_f-T_i)$

$T_f=\dfrac{-Q}{mc}+T_i$

$T_f=\dfrac{-18}{4\times 0.235}+41$

$T_f=21.85^{\circ}$

So, the final temperature of silver is 21.85 degrees Celsius.

5 0

2 years ago

In the water cycle matter moves towards the pull of gravity during ??

Bumek [7]

infiltration and evaporation

7 0

3 years ago

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

2 years ago