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Mama L [17]
3 years ago
13

Which of the following is a common ground water contaminant

Physics
1 answer:
kykrilka [37]3 years ago
5 0
The answer to the question is D:Fertilizers
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Why might a major volcanic eruption lead to cooler global temperatures? Explain your answer.
irina [24]

Answer:

Volcanic eruptions cool down  the planet

Explanation:

Volcanic eruptions actually cool the planet because the particles ejected from volcanoes shade incoming solar radiation. ... The small ash and aerosol particles decrease the amount of sunlight reaching the surface of the Earth and lower average global temperatures.

Hope this helps!!! :D

4 0
3 years ago
Two tuning forks produce sounds of wavelengths of 3.4 meters and 3.3 meters.
KengaRu [80]

Answer:

Explanation:

wavelength, λ = 3.4 m

wavelength, λ' = 3.3 m

Speed, v = 340 m/s

f = v / λ = 340 / 3.4 = 100 Hz

f' = v / λ' = 340 / 3.3 = 103.03 Hz

Frequency of beat, n = f' - f = 103.03 - 100 = 3.03 Hz

5 0
3 years ago
A company is creating a new mosquito repellent, and needs to figure out how much is needed to stop mosquito bites. They find vol
Jlenok [28]

The independent variable is the different amount of repellent each person is given. This is because it does not depend on any variable. The dependent variable is what they are measuring, which would be the number of bites on each volunteer.
I hope this helps! :)

3 0
10 months ago
What effects a materials resistance
goblinko [34]

Answer:

The type of material of which the resistor is made.

The length of the resistor.

The thickness of the resistor.

The temperature of the conductor.

8 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
3 years ago
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