Projectile motion physics question 11 please
1 answer:
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The given problem can be exemplified in the following diagram:
Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:
Replacing the values:
We may cancel out the mass:
Using the gravity constant as 9.8 meters per square second:
Solving the operations:
Therefore, the acceleration is 6.3 meters per square second.
Answer:
electricfield at (0,0,0) is Et = 2 k q / a²
Explanation:
For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity
For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation
E = k q / r²
Point (0,0,0)
We calculate for the charge -q which is at a distance R = a
E1 = k (-q) / a²
E1 = - kq / a²
As the test charge is positive in the field it goes to the left, attractive force
We calculate for the charge that is also at R = a
E2 = k q / a²
This field goes to the left, repulsive force
We find the total electric field
Et = E1 + E2
Et = kq / a² + kq / a²
Et = 2 k q / a²
Point (0,0, R)
We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a
E1 = k q / (R + a)²
E2 = kq / (R-a)²
Et = kq [1 / (R + a)² + 1 / (R-a)²]
Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}
Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}
If we use the condition that R> a we can despise in the patents "a"
(R² + a²) = R² (1+ a² / R²) ≈ R²
(R + a)² = R² (1 + a / R)² ≈ R²
(R- a)² = R² (1-a / R)² ≈ R²
Substituting in the total electric field
Et = kq {2 R²) / [R²R²]}
Et =kq 2 / R²
Explanation:
Formula depicting relation between total flux and total charge Q is as follows.
(Gauss's Law)
Putting the given values into the above formula as follows.
Q =
=
=
= -8.4 nC
Therefore, when the unknown charge is q then,
-14.0 nC + 33.0 nC + q = -8.4 nC
q = -27.4 nC
Thus, we can conclude that charge on the third object is -27.4 nC.