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tangare [24]
3 years ago
14

Carbon and hydrogen are examples of pure substances or

Physics
2 answers:
Ede4ka [16]3 years ago
7 0
The are elements on the periodic table
Softa [21]3 years ago
4 0

Answer:

Elements

Explanation:

Carbon and Hydrogen are pure substances or elements. All the atoms of carbon have same number of protons. It cannot be broken into new elements.

Contrary to elements, there are compounds which are formed by the chemical bonds between two or more elements like in water. In water, Hydrogen and oxygen elements combine. All the atoms of water do not have the same number of protons. Then, there are mixtures where elements are not chemically bound. For example, air.

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What is the force on an object that accelerates at 2 m/s/s and has a mass of 120 kg
11111nata11111 [884]

Answer:

240 Newtons

Explanatiohn:

f = m × a

f = 120 × 2

f = 240 Newtons

<h3>The force is 240 Newtons</h3>
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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.
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3 years ago
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