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Murljashka [212]
3 years ago
11

A jeweler working with a heated 47 g gold ring must lower the ring's temperature to make it safe to handle. If the ring is initi

ally at 99ºC, what mass of water at 25ºC is needed to lower the ring's temperature to 38ºC?
5.9 g
3.7 g
2.2 g
6.8 g
Physics
2 answers:
Gelneren [198K]3 years ago
6 0

Mass of gold m₁ = 47 g

Initial temperature of gold T₁ = 99 C

Specific heat of gold C₁ = 0.129 J/gC

final temperature T₂ = 38 C

Heat needed by the gold to cool down

Q =m₁ * C₁* ( T₁ - T₂)

Q = (47)(0.129)(99-38)

Q = 369.843 J

This heat will be given by the water

we need to find out mass of water m₂

and initial temperature of water is T₃ = 25 C

Specific heat of water C₂ = 4.184 J/gC

Q = m₂*C₂*(T₂ - T₃)

369.843 = m₂(4.184)(38-25)

m₂ = 6.8 g

harkovskaia [24]3 years ago
3 0
The answer is D, 6.8 g.
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Explanation:

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And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

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Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

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The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

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So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
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While an emf source supplies energy to a circuit, that energy is dissipated when the current passes through resistance. In deali
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The question is incomplete, the options are;

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P=RI^2

P=V^2/R

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Explanation:

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Answer:

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