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Gnom [1K]
3 years ago
7

A charged balloon will stick to a neutral wall. which process is involved?

Physics
1 answer:
Leokris [45]3 years ago
3 0
<span>The charged balloon will stick to a neutral wall because of the Static Electricity:
</span>
 The matter is formed by atoms and these atoms are composed of electrons, protons and neutrons (the electrons have a negative charge, the protons have a positive charge and the neutrons have no charge).

 As the balloon is charged (It gained electrons), and the charge of the same sign repel each other, when it approaches the wall, the electrons of this wall will move away, and the positive charges (protons) will remain in the nearest area to the balloon. As the charges of different signs are attracted, the balloon will be stuck to the wall.
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A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T th
N76 [4]

The mass of the ion is 5.96 X 10⁻²⁵ kg

<u>Explanation:</u>

The electrical energy given to the ion Vq will be changed into kinetic energy \frac{1}{2}mv^2

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force \frac{mv^2}{r}.

So,

Vq = \frac{1}{2}mv^2

and

Bqv = \frac{mv^2}{r}

Right from these eliminating v, we can derive

m = \frac{B^2r^2q}{2V}

On substituting the value, we get:

m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\

m = 5.96 X 10⁻²⁵ kg.

3 0
3 years ago
What is Newton's third law motion answer part a and b
Olenka [21]

Answer:

hope it helps

Explanation:

Newtons third law is that objects exert equal and opposite forces on each other.

'every action has an equal and opposite reaction'.

3 0
2 years ago
Read 2 more answers
In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi
Inessa [10]

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α  = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

8 0
3 years ago
A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien
Neporo4naja [7]

Answer:

The magnetic field is B  =  8.20 *10^{-3} \  T

Explanation:

From the question we are told that

   The  mass of the metal rod is  m  = 0.12 \ kg

    The current on the rod is  I  = 4.1 \ A

    The distance of separation(equivalent to length of the rod ) is L   = 6.3 \ m

     The coefficient of kinetic friction is \mu_k  =  0.18

      The kinetic frictional force is  F_k  = 0.212 \ N

     The constant speed is v  = 5.1 \ m/s

Generally the magnetic force on the rod is mathematically represented as  

      F  =  B * I  *   L

For  the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

        F_ k  =  B*  I  *  L

=>      B  =  \frac{F_k}{L  *  I  }

=>       B  =  \frac{0.212}{ 6.3   *  4.1   }

=>       B  =  8.20 *10^{-3} \  T

7 0
3 years ago
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1515
Talja [164]

Complete  Question

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1/5.

Answer:

The  angle is  

Explanation:

From the question we are told that

   The light emerging from second Polaroid is 1/5 the  unpolarized

Generally the intensity of light emerging from the first Polaroid is mathematically represented as

             I_1 = \frac{I_o}{ 2 }

Generally from the Malus law the intensity of light emerging from the second Polaroid  is mathematically represented

      I_2  =  I_1 cos^2 (\theta )

=>   cos^2 (\theta ) =  \frac{I_2}{I_1 }

=>   cos (\theta) =  \sqrt{ \frac{I_2}{I_1} }

From the question I_2  =  \frac{I_o}{5}

     cos (\theta) =  \sqrt{ \frac{ \frac{ I_o}{5} }{\frac{I_o}{2} } }

     cos (\theta) =  \sqrt{ \frac{2}{5} }

=>    \theta =   cos ^{-1} [\sqrt{\frac{2}{5}}  ]

=>    \theta  =  50.77^o

8 0
3 years ago
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