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Snezhnost [94]
4 years ago
6

A common demonstration involves charging a rubber balloon, which is an insulator, by rubbing it on your hair and then touching t

he balloon to a ceiling or wall, which is also an insulator. Because of the electrical attraction between the charged balloon and the neutral wall, the balloon sticks to the wall. Imagine now that we have two infinitely large, flat sheets of insulating material. One is charged, and the other is neutral. If these sheets are brought into contact, does an attractive force exist between them as there was for the balloon and the wall?
Physics
1 answer:
baherus [9]4 years ago
5 0

Answer:

here there is also an electric attraction force

Explanation:

In the insulating bodies the charges are not mobile, so when one of the body is charged periodically or gained some electrical charges.

When the body is neutral the net number of electric charges is zero. When the loaded body touches it, it repels the charges of the same sign whereby the induced load is of a different sign and the two adhered sheets are attracted.

With this explanation it is equivalent to that of the ball and the wall, so here there is also an electric attraction force

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Which of the following is an example of kinetic mechanical energy?
nata0808 [166]

Answer:

A

Explanation:

Kinetic energy must be moving. Potential energy has the ability to move but is not doing so at the moment.

A is likely the answer. But there's lots involved in that kind of motion.

B If the ball is elevated, it implies it is not moving yet. It has potential energy.

C Again, the spring is compressed. It will push something when it moves, but it is not moving yet.

D The load gun's bullet is not moving. It's still potential energy.

E. The mouse trap is set, but it is not moving. When the mouse eats the bait then it's potential energy will transform into kinetic energy.

8 0
3 years ago
The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,
madam [21]

Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

  • E = hf  
  • E = h.v/λ
  • λ = h/(mv)
  • E = (1/2)mv²

Where:

  • E = kinetic energy of the particle
  • λ = de-Broglie wavelength
  • m = mass of the particle
  • v = speed of the particle
  • h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

       \dfrac{m_\alpha}{m_p}=4

For the kinetic energies you find:

          \dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}

            \dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4

Thus:

           \dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}

          m_\alpha v_\alpha=m_pv_p

From de-Broglie equation, λ = h/(mv)  

       \dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1

5 0
4 years ago
I need homework help
KIM [24]
1) the weight of an object at Earth's surface is given by F=mg, where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is 
F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N

2) On Mars, the value of the gravitational acceleration is different:g=3.7 m/s^2. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus: 
g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg: 
g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as 
g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2

<span>6) On Earth, the gravity acceleration is </span>g=9.81 m/s^2<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
</span>F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N<span>
</span>
5 0
3 years ago
If a girl is standing still and holding a box, is she doing any work? (No)
SpyIntel [72]

Answer:

gravity

Explanation:

7 0
3 years ago
Read 2 more answers
You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

f(x,t)=Acos(kx-\omega t)

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}

k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}

T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m

b) Hence, the wave function is:

f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)

c) for x=3m you have:

f(3,t)=0.075cos(1.047*3-12.56t)

d) the speed of the medium:

\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)

you can see the velocity of the medium for example for x = 0:

v=\frac{df}{dt}=13.15cos(12.56t)

7 0
3 years ago
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