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
Answer:
1.1648×10⁻¹¹ N
Explanation:
Using
F = qvBsinФ..................... Equation 1
Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.
Note: The angle between v and B = 90°
Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°
Substitute into equation 1
F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°
F = 11.648×10⁻¹²
F = 1.1648×10⁻¹¹ N.
Answer:
k = 17043.5 N/m = 17.04 KN/m
Explanation:
First we need to find the force applied by safe pn the spring:
F = Weight of Safe
F = mg
where,
F = Force Applied by the safe on the spring = ?
m = mass of safe = 800 kg
g = 9.8 m/s²
Therefore,
F = (800 kg)(9.8 m/s²)
F = 7840 N
Now, using Hooke's Law:
F = kΔx
where,
K = Spring Constant = ?
Δx = compression = 46 cm = 0.46 m
Therefore,
7840 N = k (0.46 m)
k = 7840 N/0.46 m
<u>k = 17043.5 N/m = 17.04 KN/m</u>
Answer:you in connections too?
Explanation:
Answer: Electromagnetic radiation
Explanation:
Electromagnetic radiation is a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.
To understand it better:
This radiation is spread thanks to the electromagnetic fields produced by moving electric charges and their sources can be natural or man-made.
It should be noted that the energy of electromagnetic radiation can vary and depending on its frequency it can be useful for various situations.
