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notsponge [240]
3 years ago
8

Calculate the minimum amount of energy, in joules, required to completely melt 145 g of silver initially at 22.0°C. The melting

point of silver is at 962°C. Its specific heat capacity is 236 J/kg·K and its latent heat of fusion is 105 kJ/kg.
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

47391.8 J

Explanation:

Q = m c Δt

For melting, Δt = (962-22)°C

Heat required to attain melting point = 145 x 10 ⁻³ x 236 x 940 =

= 32166.8 J

Heat required to melt at melting point = m L , L is latent heat of fusion.

= 145 x 10⁻³ x 105 x 10³ = 15225 J

Total energy required = 47391.8 J

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For a 50 kV anode voltage, what is the maximum photon energy of the x-ray radiation?
V125BC [204]

Answer:

The energy of photon, E=8\times 10^{-15}\ J

Explanation:

It is given that,

Voltage of anode, V=50\ kV=50\times 10^3\ V=5\times 10^4\ V

We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

E=eV

e is charge of electron

E=1.6\times 10^{-19}\times 5\times 10^4

E=8\times 10^{-15}\ J

So, the maximum energy of the x- ray radiation is 8\times 10^{-15}\ J. Hence, this is the required solution.

6 0
3 years ago
SOMEONE HELP PLEASE ASAP I BELIEVE THE ANSWER IS C OR D
Serhud [2]
You are correct the answer would be C

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4 0
4 years ago
The amplitude of a lightly damped oscillator decreases by 4.2% during each cycle. What percentage of the mechanical energy of th
ra1l [238]

Answer:

V= A ω      maximum KE of object in SHM

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KE2 / KE1 = 1/2 m V2^2 / (1/2 m V1^2) = (V2 / V1)^2

KE2 / KE1 = .958^2 = .918

So KE2 = .918 KE1 and .082 = 8.2% of the energy is lost in one cycle

6 0
3 years ago
Calculate the acceleration of a turtle going from 0.3 m/s to 0.7 m/s in 30 seconds.
Hoochie [10]
Acceleration = velocity/ time
Acceleration = 0.7-0.3 /30= 0.01 m/s^2
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6 0
2 years ago
A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep
salantis [7]
The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force F_f which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
F-F_f=0
so
F=F_f

The frictional force can be rewritten as
F_f = \mu m g
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3 years ago
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