Explanation:
(a) First, we will calculate the number of moles as follows.
No. of moles = 
Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).
Putting the given values into the above formula as follows.
No. of moles =
= 
= 25 mol
According to the ideal gas equation,
PV = nRT
or, 
= 336.17 K
Hence, temperature change will be 336.17 K.
(b) The total amount of heat required for this process will be calculated as follows.
q = 
= 
= 174573.081 J/K
or, = 174.57 kJ/K (as 1 kJ = 1000 J)
Therefore, the amount of total heat required is 174.57 kJ/K.
Answer:
Explanation:
Although this isn't as overtly involved, Newton's Third Law of Motion is also important because it helps you to better design the Ferris Wheel so that the centripetal force of the wheel's center is opposed by an equal force from the objects being spun around by the wheel.
Hope this helps!!
First let us calculate for the moles of CH3OH formed:
moles CH3OH = 23 g / (32 g / mol) = 0.71875 mol
We see that there are 2 moles of H2 per mole of CH3OH, so:
moles H2 = 0.71875 mol * 2 = 1.4375 mol
Assuming ideal gas behaviour, we use the formula:
PV = nRT
V = nRT / P
V = 1.4375 mol * (62.36367 L mmHg / mol K) * (90 + 237.15
K) / 756 mm Hg
<span>V = 43.06 Liters</span>
a)
• P = F/A
P = pressure = 630 N/m^2
F = force
A = area
F = mg = 0.50 kg x 9.8 m/s^2 = 4.9 N
m= mass
g= gravity
P = F/A
A = F/P
A = 4.9 N / 630 N/m^2 = 7.778 x 10^-3 m^2
b)
• Area of a circle = pi* radius ^2
7.778 x 10^-3 m^2 = pi* radius ^2
√(7.778 x 10^-3 m^2 / pi ) = radius
radius = 0.04976 m
Answers:
a ) 7.778 x 10^-3 m^2
b) 0.04976 m
The object's speed will not change.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
