Its 1.0*10^-7M its considered a concentration because hydrogen ion is exactly equal to hydroxide ions produced by dissociation of water
Galileo Galilei is one of the key figures in the history of Science, being the first to apply the experimental-mathematical scientific method. He carried out experiments and careful observations in kinematics (his studies on the trajectory of projectiles are famous) and dynamics (it should be noted his careful experiments with inclined planes), establishing the first law of Dynamics (which Newton will later collect and refine in his Principles); and in Astronomy, with which he could unequivocally support the heliocentric theory.
His experiments were addressed by methodologies that allowed him to precisely find his mathematical calculations and to verify theories he was developing over time. His manuscripts were key to disseminate the applied method and extrapolate them to other scientific areas.
Therefore the correct answer is C.
Answer:
F = m a = m v / t where v is the change in velocity in time t
F = p / t since m v is equal to p
F = 2.2 (kg m / s) / 1.1 s = 2 kg-m / s^2 = 2 N
Or you can use the impulse equation
Answer:
D. gravity
Explanation:
Gravity keeps the atmosphere from escaping into space.
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
