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NARA [144]
3 years ago
6

What is the weight, in pounds, of a 195-kg object on jupiter?

Physics
2 answers:
Nostrana [21]3 years ago
6 0
429.901 is the weight of 195 kg on Jupiter
kirza4 [7]3 years ago
6 0

Answer:

429.9009 pounds

Explanation:

As we know that the 1 kg is related with the pound as,

1kg=2.20462pounds

Given that the weight of an object on Jupiter is,

W=195kg

Now we can convert this into the unit of pounds.

W=195kg(\frac{2.20462 pounds}{1kg})\\ W=429.9009pounds

Therefore the weight of the object on Jupiter is 429.9009 pounds.

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What is the magnitude of the velocity when the elastic potential energy is equal to the kinetic energy? (Assume that U=0 at equi
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Explanation:

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\omega =angular frequency

t=time

At any time Total Energy is the sum of kinetic Energy and Elastic potential Energy i.e. \frac{1}{2}kA^2

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\frac{1}{2}kA^2=2\times \frac{1}{2}kx^2

x=\pm \frac{A}{\sqrt{2}}

at x=\frac{A}{\sqrt{2}}

velocity is v=\frac{A\omega}{\sqrt{2}}

6 0
3 years ago
What is TRUE about a capitalist society?
Makovka662 [10]

The idea behind capitalism is that the free market of products and ideas is owned and driven by private citizens. A capitalist society is a social order in which private property rights and the free market serve as the basis of trade, distribution of goods, and development.

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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

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Thus, the woman can climb up to 64.94 meters. 

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