Answer:
the work required for the loading of second dart is 64 times greater as work required for loading the first dart.
Explanation:
k = spring constant of the spring loaded toy dart gun
x₁ = compression of spring to load the first dart = d
x₂ = compression of spring to load the second dart = 8 d
E₁ = Work required to load the first dart
E₂ = Work required to load the second dart
Work required to load the first dart is given as
E₁ = (0.5) k x₁² = (0.5) k d²
Work required to load the second dart is given as
E₂ = (0.5) k x₂² = (0.5) k (8d)² = (64) (0.5) k d²
E₂ = 64 E₁
So the work required for the loading of second dart is 64 times greater as work required for loading the first dart
Answer:
The value 
Explanation:
From the question we are told that
The volume blood ejected is 
The velocity of the blood ejected is 
The density of blood is 
The heart beat is 
The average force exerted by the blood on the wall of the aorta is mathematically represented as
=> 
=>
All three choices are correct.
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
Answer:
3.10 mole of C3H8O change in entropy is 89.54 J/K
Explanation:
Given data
mole = 3.10 moles
temperature = -89.5∘C = -89 + 273 = 183.5 K
ΔH∘fus = 5.37 kJ/mol = 5.3 ×10^3 J/mol
to find out
change in entropy
solution
we know change in entropy is ΔH∘fus / melting point
put these value so we get change in entropy that is
change in entropy 5.3 ×10^3 / 183.5
change in entropy is 28.88 J/mol-K
so we say 1 mole of C3H8O change in entropy is 28.88 J/mol-K
and for the 3.10 mole of C3H8O change in entropy is 3.10 ×28.88 J/K
3.10 mole of C3H8O change in entropy is 89.54 J/K
