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zzz [600]
3 years ago
5

What can we conclude from the attractive nature of the force between a positively charged rod and an object?

Physics
1 answer:
Paraphin [41]3 years ago
6 0

Answer:

E; The object is negatively charged

Explanation:

Here, we want to state the conclusion that can be drawn from a positively charged rod being attracted to an object.

Generally as we know, oppositely charged materials attract while the ones with same charges repel each other.

Thus, in this case, for the rod to attract the object, there must have been an opposite charge of negativity on the object

So we conclude that the reason why the rod attracted the object was because of the presence of opposing charges on both of them. And since the rod has taken the positive charge, it is only correct to state that the object is negatively charged

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The electric field direction is defined by the direction of the force felt by (select one of the following answers):A. A negativ
steposvetlana [31]

Answer:

B

Explanation:

The formula for the electric field is Force (N)/charge(Coulombs). The electric field direction is defined by the direction of the force felt by a positive charge.

4 0
3 years ago
A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
Can someone help me answer this
netineya [11]

Answer: b is sedimentary. c is metamorphic. and a is igneous.

Explanation:

6 0
3 years ago
Read 2 more answers
A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at
defon

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been  ( To / (√ (1 - (v²/c²))).  where To = 2.00 sec

8 0
3 years ago
A paper pinwheel is spinning in the wind. Which statement is correct about the forces responsible for the rotation?
VikaD [51]

Answer:

Only the perpendicular component of gravity is responsible for the rotation because wind points toward the pivot.

Explanation:

3 0
3 years ago
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