Wavelength = speed / frequency
= (350 m/s) / (440/s)
=. 0.795 meter
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
Learn more about force here:
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... low frequency and high amplitude.
Answer:
Minimum coefficient of kinetic friction between the surface and the block is 
.
Explanation:
Given:
Mass of the block = M
Spring constant = k
Distance pulled = x
According to the question:
<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed. </em>
So,
From the FBD we can say that:
⇒ Normal force, 
<em>...equation(i)</em>
⇒ Elastic potential energy, 
= 
<em> ...equation (ii)</em>
⇒ Frictional force, 
= 
<em> ...equation (iii)</em>
⇒ Plugging (i) in (iii).
⇒ 
Now,
⇒ As we know that the energy lost due to friction is equivalent to PE .
⇒ 
<em>...considering PE as</em> 
or 
.
Arranging the equation.
⇒ 
⇒ 
<em>...eliminating x from both sides.</em>
⇒ 
<em>...dividing both sides wit Mg.</em>
Minimum coefficient of kinetic friction between the surface and the block is .
