Answer:
Explanation:
Mass of block=10 kg
Applied horizontal force =F=20 N
Friction force=f=10 N
We have to find the acceleration of block.
Net force=Applied horizontal force-friction force
Where F= Horizontal force
f=Friction force
m=Mass of object
a=Acceleration of object
Hence, the acceleration of the block=
Answer
given,
change in enthalpy = 51 kJ/mole
change in activation energy = 109 kJ/mole
when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.
where as activation energy of the product and the reactant decreases.
example:
ΔH = 51 kJ/mole
E_a= 83 kJ/mole
here activation energy decrease whereas change in enthalpy remains same.
You haven't told us what the passing percentage is on the exam,
or what the passing percentage is for the semester, or any of that.
Answer: A,B, and E
Explanation: Just checked I got them right:)
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
