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VLD [36.1K]
4 years ago
11

A 2.00-kilogram object weighs 19.6 newtons on earth. if the acceleration due to gravity on mars is 3.71 meters per second2, what

is the object’s mass on mars
Physics
1 answer:
Harrizon [31]4 years ago
4 0
Recall that mass is the amount of matter present in a body. That means it's a property that is consistent regardless of the body's current location, gravity's pull on the body, etc. 

Let's not confuse mass with weight (which is a force computed as Weight = mass x acceleration). Mass will remain constant and that means that whether the object is on Earth or on Mars, its mass remains the same.  Thus, the object will still have 2.00 kg as mass on Mars. 

Answers: 2.00 kilograms
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Explanation:

5 0
3 years ago
Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature of the gas is 28.0
mel-nik [20]

Answer:

63.5 °C

Explanation:

The expression for the calculation of work done is shown below as:

w=P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

Also,

Considering the ideal gas equation as:-

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 8.314 J/ K mol

So,

V=\frac{nRT}{P}

Also, for change in volume at constant pressure, the above equation can be written as;-

\Delta V=\frac{nR\times \Delta T}{P}

So, putting in the expression of the work done, we get that:-

w=P\times \frac{nR\times \Delta T}{P}=nR\times \Delta T

Given, initial temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.0 + 273.15) K = 301.15 K

W=1770 J

n = 6 moles

So,

1770\ J=6 moles\times 8.314\ J/ Kmol \times (T_2-301.15\ K)

Thus,

T_2=301.15\ K+\frac{1770}{6\times 8.314}\ K

T_2=336.63\ K

The temperature in Celsius = 336.63-273.15 °C = 63.5 °C

<u>The final temperature is:- 63.5 °C</u>

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Answer:

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Explanation:

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6 0
3 years ago
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