Use a(t) = -32 feet per second per second as the acceleration due to gravity. (Neglect air resistance.) -With what initial veloc
2 answers:
You might be interested in
Explanation:
- Newton's first law of motion:
"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force
In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.
- Newton's second law of motion:
"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:
where F is the net force, m is the mass of the object, and a its acceleration"
In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.
"Newton's third law of motion:
"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"
Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.
Answer:
C. Pressure gradient equals gas flow over resistance.
Explanation:
As we know that pressure gradient is the driving force for the gas to flow from one point to other point
And we know that the flow rate is directly proportional to the driving force and it inversely depends on the resistance to flow
so we can say
Flow Rate =
Flow Rate =
so we can say that correct statements are as below
A. Gas flow equals pressure gradient over resistance.
B. Resistance equals pressure gradient over gas flow.
D. The amount of gas flowing in and out of the alveoli is directly proportional to the difference in pressure or pressure gradient between the external atmosphere and the alveoli.
Answer:
Mass of the climber = 69.38 kg
Explanation:
Change in length
Load, P = m x 9.81 = 9.81m
Young's modulus, Y = 0.37 x 10¹⁰ N/m²
Area
Length, L = 15 m
ΔL = 5.1 cm = 0.051 m
Substituting
Mass of the climber = 69.38 kg