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sdas [7]
3 years ago
10

Substance A has a higher heat capacity than substance B. If the same amount of heat is added to both substances, which substance

will increase in temperature more rapidly?
A. The temperature of substance B will increase more rapidly than the temperature of substance A will.
B. The temperature of substance A will increase more rapidly than the temperature of substance B will.
C. Neither substance will increase in temperature.
Physics
2 answers:
pochemuha3 years ago
7 0

The answer

<em>The temperature of substance B will increase more rapidly than the temperature of substance A will.</em>

hope this helps



igor_vitrenko [27]3 years ago
5 0

Substance-A has a higher heat capacity than substance-B has.  That means it takes more added heat to raise the temperature of Substance-A each degree.  So if the same amount of heat is added to both substances, the temperature of substance-B will increase more rapidly than the temperature of substance-A will. (choice-A)

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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

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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