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gavmur [86]
3 years ago
11

An object has a mass of 120 kg on the moon. What is the force of gravity acting on the object on the moon?

Physics
2 answers:
miss Akunina [59]3 years ago
8 0
For first question, Gravity is approximately up to 1/6 of Earth's gravity. So object's weight 1200/6=200 N. So answer shall be D.

For second question, intertia is the thing that described in the question. Answer is D.
guapka [62]3 years ago
7 0

Well first of all, if the object's mass is 120 kg, then its mass is 120 kg. 
It makes no difference where it happens to be located at the moment. 
Its mass is 120 kg. Period.  It doesn't change.

Weight = (mass) x (acceleration of gravity) .

Acceleration of gravity on the moon is 1.622 m/s² (rounded) .

Weight of the object = (120 kg) x (1.622 m/s²) = 194.6 Newtons.
                                                                   (about 43.8 pounds)

The nearest choice is  D). 196 N.

===========================================

When the bus starts moving forward, the man remains still. 
He leans back, and soon catches up.

When the bus is moving but begins to slow down, the man
continues moving forward at the same speed.  He leans forward,
until the bus stops, and then he eventually stops.

The simplest way to explain this (although it doesn't tell you much)
is to invoke 'inertia'.  'D' is the choice of the choices.

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200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.
balu736 [363]

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a. -0.01324 kJ/K,  b.  = 0.03233 kJ/K , c.  = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

<u>Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)</u>

3.54 = 0.1 . X x 203.29

<u>x = 0.1741, which is the dryness fraction of fluid at the final state.</u>

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= <u>20.329 kJ, which is much greater than 3.45 kJ</u>, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.  

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

<u>Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.</u>

<u />

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Answer:

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- Charging by contact: this is done by putting in contact a charged object with a neutral, conducting object. In this case, the charges are transferred from the charged object to the neutral object; at the end of the process, the neutral object will also have a net electric charge, so it will be also charged.

- Charging by induction: in this case, we take a charged object, and a neutral object, and we place the two objects close to each other, but without touching. Let's assume that the charged object is negatively charged: in this case, the positive charges in the neutral object are attracted towards the negative charges of the charged object, while the negative charges of the neutral object are repelled away. As a result, the positive and negative charges in the neutral object split apart. If the object is connected to the ground, then negative charges move away, so the neutral object will remain positively charged.

Therefore, the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch.

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