Democritus was the one who did not have experimental evidence to support his theory of the atom.
Answer: Option 4
<u>Explanation:
</u>
The discovery of atoms were first stated by Democritus but due to the absence of any experimental proof, his statement was not noted as significant at that time.
After this, Dalton made the specific assumptions formulating some postulates for the atomic theory with proof. Then the cathode rays tube experiments performed by Thomson lead to the formation of plum pudding models of atom.
This is followed by Rutherford’s gold foil experiment discovering the presence of nucleus inside the atoms. So, Democritus first stated but due to absence of experimental evidences, his theory of atoms were not supported at that time.
Answer with Explanation:
We are given that
Mass of box=35 kg
Coefficient of static friction between box and truck bed=0.202
Acceleration due to gravity=
a.We have to find the force by which the box accelerates forward.
Force by which box accelerates=
Force by which box accelerates=
b.We have to find the maximum acceleration can the truck have before the box slides.
Force =friction force


Hence, the truck can have maximum acceleration before the box slide=
Answer:
Pressure = 5 x 10⁶ Pa
Explanation:
Given:
Height of building = 512 m
Find:
Pressure
Computation:
P2 = P1+dgh
P2 = 1 + (1000)(9.8)(512)
P2 = 51.2 atm
Pressure = 5 x 10⁶ Pa
Answer:
The temperature of the steam during the heat rejection process is 42.5°C
Explanation:
Given the data in the question;
the maximum temperature T
in the cycle is twice the minimum absolute temperature T
in the cycle
T
= 0.5T
now, we find the efficiency of the Carnot cycle engine
η
= 1 - T
/T
η
= 1 - T
/0.5T
η
= 0.5
the efficiency of the Carnot heat engine can be expressed as;
η
= 1 - W
/Q
where W
is net work done, Q
is is the heat supplied
we substitute
0.5 = 60 / Q
Q
= 60 / 0.5
Q
= 120 kJ
Now, we apply the first law of thermodynamics to the system
W
= Q
- Q
60 = 120 - Q
Q
= 60 kJ
now, the amount of heat rejection per kg of steam is;
q
= Q
/m
we substitute
q
= 60/0.025
q
= 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q
= h
= 2400 kJ/kg
now, at h
= 2400 kJ/kg from saturated water tables;
T
= 40 + ( 45 - 40 ) (
)
T
= 40 + (5) × (0.5)
T
= 40 + 2.5
T
= 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C
Answer:
I think B or C it won't lower so I'll go with B bc warm water is hotter than regular temp water