Answer:
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
Explanation:
First, we must calculate the resultant force (
), in newtons, by vectorial sum:
![\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}](https://tex.z-dn.net/?f=%5Cvec%20F%20%3D%20%5B%28-200%5C%2CN%29%5Ccdot%20%5Ccos%2060%5E%7B%5Ccirc%7D%2B%28400%5C%2CN%29%5Ccdot%20%5Ccos%2045%5E%7B%5Ccirc%7D%2B300%5C%2CN%5D%5C%2C%5Chat%7Bi%7D%20%2B%20%5B%28200%5C%2CN%29%5Ccdot%20%5Csin%2060%5E%7B%5Ccirc%7D%20%2B%20%28400%5C%2CN%29%5Ccdot%20%5Csin%2045%5E%7B%5Ccirc%7D-100%5C%2CN%5D%5C%2C%5Chat%7Bj%7D)
(1)
Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:
Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:
Where 
is the direction of the resultant force, in sexagesimal degrees.
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
A. Coming out near the South Pole and going in near the North Pole
Answer:
A) F=-20.16×10⁹N
B) if the distance doubles, force is 4 times smaller.
Explanation:
q1=-28C
q2=5mC=0.005C
d=25cm=0.25m
Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.
Thus:
F=9×10⁹×(-28)×0.005/0.25²
F=-20.16×10⁹N
The minus sign indicates attraction.
If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.
Answer:
(A.)Nuclear fission and beta decay (electron emission)
