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d1i1m1o1n [39]
3 years ago
14

Pam rubs a balloon on her head to generate a static charge. She holds the balloon up against a wall. Which of the following desc

ribes the electric charges and forces at work if the balloon sticks to the wall?
Electrons move from Pam’s hair to the balloon and attract electrons in the wall.

Electrons move from Pam’s hair to the balloon and attract protons in the wall.

Protons move from Pam’s hair to the balloon and attract protons in the wall.

Protons in the balloon form chemical bonds with protons in the wall.
Physics
1 answer:
cestrela7 [59]3 years ago
6 0

Answer:

c

Explanation:

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Alex17521 [72]
This claim isn’t true. This claim is trying to say that once force is being applied to Newton’s Cradle then it will forever stay in motion. However, from the excerpt we learn that this isn’t possible. As one of the balls are pushed, it is set into kinetic energy, and then that ball will hit another and send it into kinetic energy as well. However, not all of the kinetic energy is kept through this process, some of the energy is lost and converted into different forms such as sound energy. Therefore, it isn’t possible for Newton’s Cradle to stay in motion forever.
3 0
3 years ago
The sound intensity level from one solo flute is 70 dB. If 10 flutists standing close together play in unison, what will the sou
Masja [62]

Answer:

80 dB

Explanation:

I_0 = Threshold intensity = 10^{-12}\ W/m^2

I = Intensity of sound

\beta = Intensity level of sound = 70 dB

Intensity level of sound is given by

\beta=10log\dfrac{I}{I_0}\\\Rightarrow 70=10log\dfrac{I}{I_0}\\\Rightarrow \dfrac{70}{10}=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{\dfrac{70}{10}}=\dfrac{I}{10^{-12}}\\\Rightarrow \dfrac{I}{10^{-12}}=10^{\dfrac{70}{10}}\\\Rightarrow I=10^{7}\times 10^{-12}\\\Rightarrow I=10^{-5}\ W/m^2

If there are 10 flutes I=10\times 10^{-5}\ W/m^2

\beta=10log\dfrac{10\times 10^{-5}}{10^{-12}}\\\Rightarrow \beta=10log10^8\\\Rightarrow \beta=10\times 8\\\Rightarrow \beta=80\ dB

The sound intensity level is 80 dB

5 0
3 years ago
Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache
HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

8 0
3 years ago
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Answer:

H2O is also known as for water

5 0
2 years ago
Design a solution that can monitor and minimize the melting of sea ice caused by human activity
dedylja [7]

Answer:

CO2 emissions from fossil fuel burning should be minimized at all cost. The CO2 are gotten when the carbons from hydrocarbons react with air(oxygen). This gas erodes the ozone layer which makes the melting of ice caps faster due to increased amount of heat radiations on the earth. This is the only best and permanent solution to the reduction of the amount of heat rays on the earth which is a global problem.

Objects which reflects back the sunrays could also be inserted into the sea to prevent the melting of the ice caps.

7 0
3 years ago
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