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Tasya [4]
3 years ago
14

This problem has been solved! See the answer A 6.0 kg object, initially at rest in free space, "explodes" into three segments of

equal mass. Two of these segments are observed to be moving with equal speeds of 20 m/s with an angle of 60 degrees between their directions of motion. How much kinetic energy is released in this explosion?
Physics
1 answer:
Monica [59]3 years ago
5 0

Answer:

Explanation:

mass of each part = 6 / 3 = 2 kg .

momentum of each of given part = 2 x 20 = 40 kg m/s

Two momentum of 40 kg m/s , acting at angle 60 degree .

Resultant momentum =  2 x 40 cos 30 = 69.28 kg m/s

The third mass will have equal and opposite momentum to this momentum , following law of conservation of momentum .

If its velocity be v .

2 x v = 69.28

v = 34.64 m /s

Third mass will have velocity of 34.64 m /s

Total kinetic energy of all three mass

KE = 1/2 x 2 ( 20² + 20² + 34.64² )

= 400 + 400 + 1199.92

= 1999.93 J .

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Answer:

t = 106π / 30*2.1

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w_{i} = 1.06*10^{2}    => 106

    => 106 x 2π/60

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∝ = -2.1 rad/sec²

w_{f} => 0

w_{f} = w_{i} + ∝t

∴ (w_{f} - w_{i}) / ∝ = t

t = 106π / 30*2.1

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One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th
lozanna [386]

Answer:

k=105.0359\times 10^4\,W.m^{-1}.K^{-1}

Explanation:

Given:

temperature at the hotter end, T_H=100^{\circ}C

temperature at the cooler end, T_C=0^{\circ}C

length of rod through which the heat travels, dx=0.7\,m

cross-sectional area of rod, A=1.1\times 10^{-4}\,cm^2

mass of ice melted at zero degree Celsius, m=8.7\times 10^{-3}\,kg

time taken for the melting of ice, t=15\times60=900\,s

thermal conductivity k=?

By Fourier's Law of conduction we have:

\dot{Q}=k.A.\frac{dT}{dx}......................................(1)

where:

\dot{Q}=rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice,  L = 334000\,J.kg^{-1}

Q=m.L

Q=8.7\times 10^{-3}\times 334000

Q=2905.8\,J

Now the heat rate:

\dot{Q}=\frac{Q}{t}

\dot{Q}=\frac{2905.8}{900}

\dot{Q}=3.2287\,W

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3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}

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