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Brilliant_brown [7]
3 years ago
9

The temperature of a 700.96 gram piece of metal falls 120⁰C and in the process releases 2001 Joules of energy. What is the speci

fic heat of the piece of metal?
Physics
1 answer:
olga nikolaevna [1]3 years ago
8 0

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C  - 100°C   = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

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20. Which statement is false? Rewrite it so that it is true. a. Fusion involves the combination of two smaller atoms into a larg
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This statement is false. In fact, both fission and fusion are processes which release very large amounts of energy. The statement can be rewritten as it is true as follows:

"Fission and fusion are two processes that release very large amounts of energy."

Fission occurs when a large nucleus break apart, splitting into smaller nuclei, while fusion occurs when two light nuclei combine together into a larger nucleus. In both cases, the mass of the reactants is larger than the mass of the final products, so some of the mass has been converted into energy, according to Einstein's equation:

E = \Delta m c^2

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3 0
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At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet w
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Explanation:

Formula for the electric field due to the infinite sheet of charge is as follows.

               E = \frac{\sigma}{2 \epsilon_{o}}

where,   \sigma = surface charge density

Now, formula for electric force acting on the proton is as follows.

             F = eE

where,    e = charge of the proton

According to the Newton's second law of motion, the net force acting on the proton is as follows.

                       F = ma

                 a = \frac{eE}{m}

                    = \frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}

                    = \frac{e \sigma}{2m \epsilon_{o}}

According to the kinematic equation, speed of the proton in perpendicular direction is as follows.

              v_{f} = v_{i} + at

                     = (0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t

                     = \frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}

                     = 683.974 m/s

Hence, total speed of the proton is as follows.

                v' = \sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}

                    = \sqrt{921600 + 467820.43}

                    = \sqrt{1389420.43}

                    = 1178.73 m/s

Therefore, we can conclude that speed of the proton is 1178.73 m/s.

3 0
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