Use Newton's second law and the free body diagram to determine the net force and acceleration of an object. In this unit, the forces acting on the object were always directed in one dimension.
The object may have been subjected to both horizontal and vertical forces but there was no single force directed both horizontally and vertically. Moreover, when free-body diagram analysis was performed, the net force was either horizontal or vertical, never both horizontal and vertical.
Times have changed and we are ready for situations involving two-dimensional forces. In this unit, we explore the effects of forces acting at an angle to the horizontal. This makes the force act in two dimensions, horizontal and vertical. In such situations, as always in situations involving one-dimensional network forces, Newton's second law applies.
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Answer:
-30 N/C
Explanation:
Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m
Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V
Since E = -ΔV/Δx
substituting the values of the variables into the equation, we have
E = -ΔV/Δx
E = -0.30 V/0.01 m
E = -30 V/m
Since 1 V/m = 1 N/C.
E = -30 N/C
So, the average electric field is -30 N/C
Answer:
Frequency, 
Explanation:
Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?
It is given that,
Wavelength of a visible red light is, 
Speed of light is, 
We need to find the frequency of visible red light. It can be calculated using below relation.
So, the frequency of visible red light is 
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