Work done by me for using force of 4 N to push an object 3 m away is 12 J.
<u>Explanation:</u>
Work done on any object is defined as the force required to displace that object from its original position. In other terms, the work done is the measure of force utilized for moving an object. So mathematically it is represented as the product of force with the displacement of the object due to that force.
Then, here it is given that force of about 4 N is used to displace the object to 3m. Thus,
So, 12 J of work is done by me to push an object 3 m away with the force of 4 N.
1) Troposphere is the nearest layer to the earth.
2) Range from 5 to 11 miles in thickness.
3) Contains almost all of the atmospheric water vapor.
Answer:
V₂ = 1500 Liters ( 2 sig. figs.)
Explanation:
Given the following gas law variables:
P₁ = 110.0KPa P₂ = 25KPa
V₁ = 410 Liters V₂ = ?
T₁ = 17°C ( = 290K) T₂ = -27°C ( = 248K)
P₁V₁/₁T₁ = P₂V₂/T₂ => V₂ = V₁(P₁/P₂)(T₂/T₂)
V₂ = 410L(110.0KPa/25KPa)(248k/290K) = 1542 L (calc. ans.)
V₂ = 1500 Liters ( 2 sig. figs.)
Answer:
Explanation:
Hello!
In this case, according to the given chemical reaction, it is possible to calculate the enthalpy change of the reaction via:
Since the enthalpy of formation of oxygen is 0. Thus, given the enthalpies of formation of gaseous carbon dioxide and water, we obtain:
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