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 final velocity becomes 31.48 m/s
<u>Explanation:</u>
Given:
Initial velocity, u = 33 m/s
Height, h = 5m
Final velocity, v = ?
According to Newton's law:
v² - u² = 2gh
where,
g is the acceleration due to gravity and
g = 9.8 m/s²
On substituting the values we get:

Therefore, the final velocity becomes 31.48 m/s
Using the given equation you get:
E = 1.99x10^-25 / 9.0x10^-6
Divide 1.99 by 9.0: 1.99/9.0 = 0.22
For the scientific notation, when dividing subtract the two exponents:
25 -6 = 19
So you now have 0.22 x 10^-19
Now you need to change the 0.22 to be in scientific notation form:
2.2 x 10^-20
The answer is B.
Answer:
A. 50 m/s
Explanation:
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 4 s
Find: v
v = at + v₀
v = (10 m/s²) (4 s) + 0 m/s
v = 40 m/s
In the x direction, the velocity is constant at 30 m/s.
The overall speed is:
v² = (30 m/s)² + (40 m/s)²
v = 50 m/s