Answer:
V₂ = 1.5 m/s
Explanation:
given,
speed of the first piece = 6 m/s
speed of the third piece = 3 m/s
speed of the second fragment = ?
mass ratios = 1 : 4 : 2
fragment break fly off = 120°
α = β = γ = 120°
sin α = sin β = sin γ = 0.866
using lammi's theorem
A,B and C is momentum of the fragments
4 x V₂ = 2 x 3
V₂ = 1.5 m/s
Answer:
q₁= +0.5nC
Explanation:
Theory of electrical forces
Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
To solve this problem we apply Coulomb's law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
o solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters
Data:
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Data
q₃=+5.00 nC =+5* 10⁻⁹ C
q₂= -2.00 nC =-2* 10⁻⁹ C
d₂= 5.00 cm= 5*10⁻² m
d₁= 2.50 cm= 2.5*10⁻² m
k = 8.99*10⁹ N*m²/C²
Calculation of magnitude and sign of q1
Fn₃=0 : net force on q3 equals zero
F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.
F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.
We propose the algebraic sum of the forces on q₃
F₂₃ - F₁₃=0
We eliminate k*q₃ of the equation
q₁= +0.5*10⁻⁹ C
q₁= +0.5nC
Answer:
1.23917 m
0.14323 s
Explanation:
v = Speed of sound in dry air at 20 °C = 343.5 m/s
f = Frequency of note C# = 277.2 /s = 277.2 Hz
λ = Wavelength
Wavelength = 1.23917 m
Distance the wave needs to travel is 49.2 m
Time = Distance / Speed
Time taken for the sound to travel across the concert hall is 0.14323 s