The net force on q2 will be 1.35 N
A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe force. Being a vector quantity, a force has both magnitude and direction.
Given Particles q1, q2, and q3 are in a straight line. Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C, and q3 = -2.50 x 10-6 C. Particles q₁ and q2 are separated by 0.500 m. Particles q2 and q3 are separated by 0.250 m.
We have to find the net force on q2
At first we will find Force due to q1
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.5²
F = 450 × 10⁻³
F₁ = 0.45 N (+)
Now we will find Force due to q2
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.25²
F = 1800 × 10⁻³
F₂ = 1.8 N (-)
So net force (F) will be
F = F₂ - F₁
F = 1.8 - 0.45
F = 1.35 N
Hence the net force on q2 will be 1.35 N
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Answer: 30
Explanation: it is 30 meters per seconds
Answer:
1.2×10⁶ Hz
Explanation:
Applying,
v = λf............. Equation 1
Where v = speed of wave, λ = wavelength, f = frequency.
make f the subject of the equation
f = v/λ............... Equation 2
From the question,
Given: v = 3.0×10⁸ m/s, λ = 2.5×10² m
Substitute these values into equation 2
f = 3.0×10⁸/2.5×10²
f = 1.2×10⁶ Hz
Answer:
Part A:
Part B:
Explanation:
Part A:
To calculate the number of free electrons n we use the following formula::
n=1.5N-Au
Where N-Au is number of gold atoms per cubic meter
So:
Part B:
n is calculated above which is 8.85*10^{28}m^{-3}
Charge on electron=1.602*10^{-19}
Elec- Conductivity= 4.3*10^{7}
