Answer:
According to newton's second law of motionF=ma Data:-F=3200kgm/sec² or N ,a=2m/sec² ,m=? solution :-F=ma here we have to find m so m=F/a ,m=3200/2=1600kg
All three have the same current, so that is not a factor. Wattage (power) is E*I or i^2 R. The higher the resistance, the more power dissipated. The answer is R3 because it has the highest resistance.
R3 <<<< ===== answer.
The problem is solved and the questions are answered below.
Explanation:
a. To calculate the speed of the 0.66 kg ball just before the collision
V₀ + K₀ = V₁ + K₁
= mgh₀ = 1/2 mv₁²
where, h= r - r cosθ
V = 
V = 2.42 m/s
b. Calculate the speed of the 0.22 kg ball immediately after the collision
y = y₀ + Vy₀t - 1/2 gt²
0 = 1.2 - 1/2 gt²
t = 0.495 s
x = x₀ + Vx₀t
1.4 = 0 + vx₀ (0.495)
Vx₀ = 2.83 m/s
C. To Calculate the speed of the 0.66 kg ball immediately after the collision
m₁ v₁ = m₁ v₃ + m₂ v₄
(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)
V₃ = 1.48 m/s
D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.
E. To Calculate the height to which the 0.66 kg ball rises after the collision
V₀ + k₀ = V₁ + k₁
1/2 mv₀² = mgh₁
h₁ = v₀²/2 g
= 0.112 m
F. Based on your data, No the collision is not elastic.
Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²
= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²
= - 0.329 J
Hence, kinetic energy is not conserved.
Answer:

Explanation:
Where E is the magnitude of electric field...
k is called Columb's Constant. It has a value of 8.99 x 109 N m2/C2.
Qs is the magnitude of the source charge...
and r is the magnitude of distance between source and target...
(When electron comes to rest Δt the magnitude of Electric field E become zero momentarily but later achieves the maximum value...)
Answer:
See below
Explanation:
a) Spring force at release = k * d = 132 N/m * .120 m = 15.84 N
b) F = ma
15.84 = (.660 kg)(a) a = 24 m/s^2
c) Toward the left ....the object is accelerated to the left