Answer:
Fnet = F√2
Fnet = kq²/r² √2
Explanation:
A exerts a force F on B, and C exerts an equal force F on B perpendicular to that. The net force can be found with Pythagorean theorem:
Fnet = √(F² + F²)
Fnet = F√2
The force between two charges particles is:
F = k q₁ q₂ / r²
where
k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.
If we say the charge of each particle is q, then:
F = kq²/r²
Substituting:
Fnet = kq²/r² √2
Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = ![\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%261%261%5C%5C1%260%265%5Cend%7Barray%7D%5Cright%5D)
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
Answer:
915 Hz
Explanation:
The observed frequency from a sound source is given as
f₀ = f [(v + v₀)/(v+vₛ)]
where
f₀ = observed frequency of the sound by the observer = ?
f = actual frequency of the sound wave = 983 Hz
v = actual velocity of the sound waves = 343 m/s
vₛ = velocity of the source of the sound waves = 55.9 m/s
v₀ = velocity of the observer = 28.4 m/s
f₀ = 983 [(343+28.4)/(343+55.9)]
f₀ = 915.2 Hz = 915 Hz