Electric field strength = resistivity of copper x current density
where
p= 1.72 x 10^-8 <span>ohm meter
diameter = 2.05mm=.00205 m
current = 2.75 A
</span>get first the current density:
current density = current/ cross section area
find the cross section area
cross section area = pi.(d/2)^2;
cross section = 3.3 006x10-6 m^2
substitute the values
current density = 2.75A/3.3006x 10-6m^2
current density=35.55 x1 0^2 A/m^2
Electric field stregnth =1.72 x 10^-8 ohm meter x 35.55 x10^2 A/m^2
Electric field stregnth= 46.415 Volts/m
The electric field strength of copper is 46.415 V/m.
Answer:
<em>The initial speed of the sprinter was 2.2 m/s</em>
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
The following relation applies:

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.
The sprinter speeds up from an unknown initial speed to vf=3 m/s in t=2 seconds with an acceleration of
.
To find the initial speed, we solve the equation for vo:

Substituting the values:



The initial speed of the sprinter was 2.2 m/s
Answer:
An object in equilibrium has a net force of zero
Static equilibrium describes an object at rest having equal and balanced forces acting upon it.
Dynamic equilibrium describes an object in motion having equal and balanced forces acting upon it.
Explanation:
An object is said to be in equilibrium when a net force of zero is acting on it. When this condition occurs, the object will have zero acceleration, according to Newton's second law:

where F is the net force, m the mass of the object, a the acceleration. Since F=0, then a=0. As a result, we have two possible situations:
- If the object was at rest, then it will keep its state of rest. In this case, we talk about static equilibrium.
- If the object was moving, it will keep moving with constant velocity. In this case, we talk about dynamic equilibrium.
It would mean that you could not know the precise volume of the sand. Only the volume of the sand plus the water that was making it damp.
In the experiments listed, the effects are easy to deduce by understanding that the water in the sand adds volume to the 'sample' being measured.
So in the case of calculating air space you would calculate <em>less</em> air space.
70x3=210 mph... its that easy