Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = 
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = 
(|r₂₁|)² = 148.25
= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.
Let:
b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°
We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.
Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.
In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
Answer:
Negative
Explanation:
Observe that the object below moves in the positive direction with a changing velocity. An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).
Answer:
An object can have a displacement in the absence of any external force acting on it
Explanation:
When a object moves with a constant velocity (v), then it gets displaced in the direction of motion but the net external force experienced by the object is zero.
F external =ma
If object moves with constant velocity, acceleration is zero.
Since, a=0 ⟹F external =0
Using s=ut+ 1/2 at ^2
⟹ Displacement s=ut (∵a=0)
Hence, an object can have a displacement in the absence of any external force acting on it
Hope this helped you:)
