Answer:
The straight line that is obtained, intercept it on the y-axis and the value of displacement will obtained.
Explanation:
Your potential energy and mass don't tell what your weight is.
If I walk up from the first floor to the second floor, my weight hasn't
changed even though my potential energy has increased.
The resultant force on the positive charge is mathematically given as
X=40N
<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>
Question Parameters:
Three-point charges, two positive and one negative, each having a magnitude of 20
Generally, the -ve charge is mathematically given as
Q+=X
Therefore
X=40N
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M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore;
(1060× 16) + (1830 ×0) = (1060 +1830) V
16960 = 2890 V
V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
