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sdas [7]
3 years ago
13

At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the sp

eed (in m/s) of the body at this instant?m/s(b)What is the mass (in kg) of the body at this instant?kg
Physics
1 answer:
motikmotik3 years ago
5 0

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

K = \frac{1}{2} mv^2

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51  = 1.07 kg

The mass of the body at that instant is 1.07 kg

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Recall the definition of the cross product with respect to the unit vectors:

i × i = j × j = k × k = 0

i × j = k

j × k = i

k × i = j

and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)

Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have

a × b = (8i + j - 2k) × (5i - 3j + k)

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a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)

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3 years ago
Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
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Answer:

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b.0.1483Btu/s.R

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h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

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b. Heat gained by the water is equal to the heat lost by the condensing steam.

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\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

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Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

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