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polet [3.4K]
3 years ago
15

The city is thinking about starting a program to add a new chemical to the public pool. Before they do, they want to find out if

there are any health risks associated with this type of chemical. Which group would be the best to advise them on this topic?
a group of concerned citizens who oppose the program
the company that manufactures the chemical
the local pest pool companies who would be hired to take care of the pool
the government agency that regulates these types of chemicals
Physics
1 answer:
Kitty [74]3 years ago
4 0
The best and most correct answer among the choices provided by the question is the fourth choice.


The best people for advising is <span>the government agency that regulates these types of chemicals.</span>
I hope my answer has come to your help. God bless and have a nice day ahead!
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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
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A 1000.kg car moving at north at 100. km/hr brakes to a stop in 50. m within a uniform acceleration. what are magnitude and dire
Marianna [84]

Answer:

Force=7.71KN in the opposite direction

Explanation:

Given the mass of the car(M)=1000Kg

The initial speed of the car(u)=100km/hr

we know that 1km/hr=5/18 m/sec

100km/hr=100*5/18 m/sec=27.77m/sec

the distance travelled before it stops (s)=50m

let the acceleration of the car be a

The final velocity of the car is 0.

we know that  v^{2}-u^{2}=2as

0^{2}-27.77^{2}=2(-a)(50)

a=7.712m/sec^{2}

We know that F=Ma

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6 0
3 years ago
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
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Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

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3 years ago
Consider a magnetic force acting on an electric charge in a uniform magnetic field. Which of the following statements are true?
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Answer:

Pushing magma up through the interior of earth

Explanation:

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