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loris [4]
3 years ago
13

When resting, a person has a metabolic rate of about 3.0 105 joules per hour. The person is submerged neck-deep into a tub conta

ining 1.21 103 kg of water at 19.30° C. If the heat from the person goes only into the water, find the water temperature after half an hour.
Physics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

The temperature after half an hour is 19.3002^{\circ}

Solution:

As per the question;

Metabolic rate of the person is 3.0105 J/h

Temperature, T = 19.30^{\circ}

Mass of the water, m_{w} = 1.2103 kg

Time duration, t = 0.5 h = 30 min = 180 s

Now,

Heat, Q = ms\Delta t

Thus heat transfer in half an hour:

Q = 3.0105\times 0.5 = 1.505 J

Now, the temperature of water after half an hour, T' is given by:

Q = m_{w}s\Delta T = ms(T' - T)

where

s = 4186 J

1.505 = 1.21103\times 4186\times (T' - 19.30^{\circ})

T' = 19.3002 ^{\circ}

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Anna007 [38]

Answer:

w_cycle = 13.68 KJ / cycle

Explanation:

Given:

- Steady power out-put W_out = 15.5 hp

- System executes 50 cycles/min

Find:

Determine the amount of work that is delivered during each cycle in kilo-joules

Solution:

- First we will convert the steady output into KW as follows:

                                 W_out = 15.5 hp

The conversion factor is 1.36 KW per hp.

                                 W_out = 15.5 [hp] * [KW] / 1.36 [hp]

                                 W_out = 11.4 [KW] or [KJ/s]

- Given that there are 50 cycles in 60 sec. We will use direct proportionality to calculate the number of cycles per second:

                       50 cycles   -----------> 60 s

                          x cycles   -----------> 1 s

                                 x = 50/60 = 0.8333 cycles /s

- Next we will compute the amount of work done per cycle:

                                 w_cycle = W_out / x

                                 w_cycle = 11.4 [KJ/s] * [s] / 0.8333 [cycles]

                                 w_cycle = 13.68 KJ / cycle

7 0
3 years ago
A car from rest moves a distance s(m) in t second, where s=3t³+t²/4 determine(i)the initial acceleration of the car (ii)the acce
Sunny_sXe [5.5K]

Answer:

0.5 , 54.5

Explanation:

for acceleration we should derivate the equation 2 times

x=3t³+t²/4

v=9t²+t/2

a=18t+1/2

a(0)=0.5

a(3)=54.5

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3 years ago
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.
SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

  • Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
  • For the first collission, only mass 1 is moving before it, so we can write the following equation:

       p_{i} = p_{f} = m*v_{o}    (1)

  • Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

       p_{f1} = 2*m*v_{1}    (2)

  • From (1) and (2) we get:
  • v₁ = v₀/2  (3)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

       p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2}  = m*v (4)

  • Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

        p_{2} = 3*m*v_{2}  (5)

  • From (4) and (5) we get:
  • v₂ = v₀/3  (6)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

      p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3}  = m*v (7)

  • Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

       p_{3} = 4*m*v_{3}  (8)

  • From (7) and (8) we get:
  • v₃ = v₀/4
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Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small
Vaselesa [24]

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

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Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

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3 years ago
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