Answer:
The ballon will brust at
<em>Pmax = 518 Torr ≈ 0.687 Atm </em>
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Explanation:
Hello!
To solve this problem we are going to use the ideal gass law
PV = nRT
Where n (number of moles) and R are constants (in the present case)
Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.
--- (*)
Our initial state is:
P1 = 754 torr
V1 = 3.1 L
T1 = 294 K
If we consider the final state at which the ballon will explode, then:
P2 = Pmax
V2 = Vmax
T2 = 273 K
We also know that the maximum surface area is: 1257 cm^2
If we consider a spherical ballon, we can obtain the maximum radius:
Rmax = 10.001 cm
Therefore, the max volume will be:
Vmax = 4 190.05 cm^3 = 4.19 L
Now, from (*)
Therefore:
Pmax= P1 * (0.687)
That is:
Pmax = 518 Torr
1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.
<u>Explanation</u>:
We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the 
and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:
Better understood from numerical example as given:
If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?
This can be solved as follows:
It shows that man A will have more K.E.
Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.
Answer:
Tuesday bc instead of running he/she was walking bc he/she might not have as much energy
Explanation:
Answer:
The genetic makeup in a organism.
Explanation:
Answer:
33 N
Explanation:
v = Velocity of fluid = 8+2 = 10 m/s
= Density of fluid = 1.2 kg/m³
C = Coefficient of drag = 1.1
A = Cross sectional area = 0.5 m²
Drag force is given by
The drag force on the athlete is 33 N
