<span>The purpose of pumping is to increase overall velocity. The person drops down into a crouch while traversing the more-or-less flat bottom of the U-shaped pipe or bowl. Then, as he enters the sloped part of the ramp or bowl, called the transition, he straightens his legs and rises up. By raising his center of mass just at the beginning of the arc, the person gains energy and thereby increases his speed.</span>
Answer:
<em>No, a rigid body cannot experience any acceleration when the resultant force acting on the body is zero.</em>
Explanation:
If the net force on a body is zero, then it means that all the forces acting on the body are balanced and cancel out one another. This sate of equilibrium can be static equilibrium (like that of a rigid body), or dynamic equilibrium (that of a car moving with constant velocity)
For a body under this type of equilibrium,
ΣF = 0 ...1
where ΣF is the resultant force (total effective force due to all the forces acting on the body)
For a body to accelerate, there must be a force acting on it. The acceleration of a body is proportional to the force applied, for a constant mass of the body. The relationship between the net force and mass is given as
ΣF = ma ...2
where m is the mass of the body
a is the acceleration of the body
Substituting equation 2 into equation 1, we have
0 = ma
therefore,
a = 0
this means that<em> if the resultant force acting on a rigid body is zero, then there won't be any force available to produce acceleration on the body.</em>
<em></em>
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm
We will have the following:
So, the force is approximately 1.85*10^-6 N.
The electric field between plates is 4000V/m.
An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.
The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.
The voltage between points A and B is
V=E.d
E =V/d (uniform E- field only)
where d is the distance from A to B, or the distance between the plates.
Given:
distance d = 3 cm
voltage V = 120 V
Electric field E = V/d
E = 120 V / 3cm
E = 40 V / 1 cm [ 1 cm = 1/100 m ]
E = 4000 V/m.
Learn more about Electric field here:
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