Average speed is the total distance divided by the
1 answer:
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Answer:
11m/s
Explanation:
Given parameters:
Mass of the lead shot = 8kg
Energy of the shot = 484J
Unknown:
Velocity of the shot = ?
Solution:
To solve this problem, we apply the kinetic energy formula;
K.E = m v²
m is the mass
v is the unknown
Now, insert the parameters and solve;
484 = x 8 x v²
484 = 4v²
v² = 121
v = 11m/s
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
