How do i turn 36 m/s into mph

The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.

6 0

2 years ago

A 175 g ball on the end of a 0.5 m light string is revolving uniformly at 2 rev/s. what is the ball's acceleration and string's

BARSIC [14]

Hope it cleared your doubt.

3 0

3 years ago

A block of mass 5 kg is placed on a rough table having a coefficient of static friction μs = 0.2.What is the force required to m

just olya [345]

Answer:

<em>We need to (at least) apply a force of 9.8 N to move the block</em>

Explanation:

<u>Second Newton's Law</u>

If a net force $F_n$ different from zero is applied to an object of mass m, then it will move at an acceleration a, given by

$F_n=ma$

If we apply a force F to an object placed on a rough surface, the only way to make it move is to beat the friction force which is given by

$F_r=\mu F_N$

Where $\mu$ is the static friction coefficient and $F_N$ is the normal force exerted by the table to the object. Since there is no motion in the vertical direction the normal force equals the weight of the object:

$F_N=mg=5\ kg\ 9.8\ m/s^2=49\ N$

The friction force is

$F_r=0.2 (49)=9.8\ N$

Thus, we need to (at least) apply a force of 9.8 N to move the block

5 0

4 years ago