Newton’s Third Law of Motion states that for every action there is an equal and opposite reaction. So look for a scenario in which something had force applied upon it and the reaction is a force in the opposite direction of the same size.
The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.
The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N
The other free body diagram is around the joint of the three strings.
In this case, you can do the horizontal forces equilibrium equation as:
T1* cos(60) - T2*cos(40) = 0
And the vertical forces equilibrium equation:
Ti sin(60) + T2 sin(40) = T3 = 325 N
Then you have two equations with two unknown variables, T1 and T2
0.5 T1 - 0.766 T2 = 0
0.866 T1 + 0.643T2 = 325
When you solve it you get, T1 = 252.8 N and T2 = 165 N
Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N
Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C
Answer:
B A new compound is formed
Answer:
Explanation:
A proton and electron are moving in the positive x direction, this shows that their velocity will be in the positive x direction
V = v•i
Magnetic field Is the positive z direction
B = B•k
A. For proton.
Proton has a positive charge of q
Direction of force on proton
Force is given as
F = q(v×B)
F = q( v•i × B•k)
F = qvB (i×k)
From vectors i×k = -j
F = -qvB •j
Then, for the positive charge, the force will act in the negative direction of the y-axis
B. For electron
Electron has a negative of -q
Direction of force on proton
Force is given as
F = q(v×B)
F = -q( v•i × B•k)
F = -qvB (i×k)
From vectors i×k = -j
F = --qvB •j
F = qvB •j
Then, for the negative charge, the force will act in the positive direction of the y-axis
