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Alexeev081 [22]
3 years ago
10

When people use plastic combs on their hair, the combs become negatively charged. Which statements about this situation are true

?
The comb loses electrons.

The comb gains electrons.

The hair loses electrons.

The hair gains protons.

The hair loses protons.


those above are the choices
Physics
2 answers:
Ede4ka [16]3 years ago
5 0

Answer:The comb gains electrons.

The hair loses electrons

Explanation:

Electron is negatively charged with a small mass.

Since electrons are negatively charged, they will move since the comb has no charge initially when it comes in contact with the air it gains electrons from it and the hair looses electrons

Vadim26 [7]3 years ago
3 0
The answer would be:
The comb gains electrons.
The hair loses electrons.

In chemistry, the charge depends on the electrons and protons. The electron will give negative charge and proton will give the positive charge. Proton is located in the nucleus of the atoms so it won't easily move like electron which located in the orbit in the atoms perimeter. So, ignore the option with the proton.
If the combs become negatively charged, that means it gain some electron. Since something gain electron, that means another thing is losing an electron. That electron comes from the hair.
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Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

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water flows horizontally from a hose that is 3m above the ground. the water lands 7.2m to the right of the hose . how long does
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Answer:

0.782 s

Explanation:

The water flows horizontally from the hose, so its initial vertical velocity is 0.

Given:

y₀ = 3 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 3 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 0.782 s

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Answer:

b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface

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Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.

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A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert
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(a) -0.211 m

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the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

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The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

\Delta U = mg \Delta y

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

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So, the work done by gravity is

W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J

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(c) Zero

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W=Fd cos \theta

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F is the magnitude of the force

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\theta is the angle between the direction of the force and the displacement

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\theta=90^{\circ}, cos \theta = 0

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Mg -  5Ma/8 =  Ma

Mg= 5Ma/8 +  Ma = 13Ma / 8

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The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together

The size of the net balance of all external forces acting on that item is directly proportional to the magnitude of this net resultant force; the magnitude of that object's mass, depending on the materials from which it is built, is inversely related to its mass.

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