Answer:
Action-at-a-Distance Forces. Frictional Force. Gravitational Force. Tension Force ... The force of gravity on earth is always equal to the weight of the object as ... The friction force is the force exerted by a surface as an object moves across it or ... The force of air resistance is often observed to oppose the motion of an object
Explanation:
Answer:
Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
Explanation:
Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
In a series circuit the total current is the same throughout resistors and so:
The voltage is distributed throughout the resistors and so:
and the total resistance can be calculated by adding up the resistors resistance:
First thing is to calculate the total resistance and so:
And by Omh's law V=IR we have:
And so the total current of the circuit is 1.2 amps i.e. 1.2 A.
A process known as fixation<span>. the majority of nitrogen is fixed by </span>bacteria<span>, most of which are </span>symbiotic<span> with plants</span>
<span />
Answer:
Lens at a distance = 7.5 cm
Lens at a distance = 6.86 cm (Approx)
Explanation:
Given:
Object distance u = 12 cm
a) Focal length = 20 cm
b) Focal length = 16 cm
Computation:
a. 1/v = 1/u + 1/f
1/v = 1/20 + 1/12
v = 7.5 cm
Lens at a distance = 7.5 cm
b. 1/v = 1/u + 1/f
1/v = 1/16 + 1/12
v = 6.86 cm (Approx)
Lens at a distance = 6.86 cm (Approx)
