A boy is pulling his two sisters on a sled.
1 answer:
Answer:
Explanation:
The oly way we can figure this out is if the boy is pulling the sled at a constant velocity. If not, we need a value for acceleration, and you don't have that here. If the boy is pulling the sled at a constant velocity, then the value for acceleration is 0, making this a really simple problem. I'm going with that, since there is no way to answer you otherwise. If velocity is not constant, please either repost the question or put it in the notes section under the question as it stands. If acceleration is 0, then
F - f = ma becomes
F - f = m(0) which is
F - f = 0 and
F = f which says that the applied force is the same as the frictional force. We need then to find the frictional force, which has an equation of
f = μ where normal force is the same as the weight of the 2 girls. We will find that, then:
Each girl's mass is different so the normal force/weight equation is
w = (30.0)(9.8) + (40.0)(9.8) to get
w = 290 + 390 and
w = 680. Plug that into the frictional force equation:
f = (.120)(680) so
f = 82N
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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
