الملك - الإتاوات - المحاكم - المجلس الملكي - القانون الروماني
Answer:

Explanation:
Given data:
weigh (head+arms + head) w_1 = 438 N
centre of gravity y_1= 1.28 m
weigh (upper leg) w_2 = 144 N
Center of gravity y_2 = 0.760 m
weigh ( lower leg + feet) = 87 N
centre of gravity = y_3 = 0.250 m
location of center of gravity 


Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
Answer:
M2 = 278.06 kg
Explanation:
We calculate the weight of M1
W=m*g
Where
m: mass (kg)
g: acceleration due to gravity (m/s²)
W₁=288* 9.8= 2822.4 N
Look at the attached graphic
We calculate the x-y components of the weight :
W₁x= 2822.4*sin41° N =1851.66 N
W₁y= 2822.4 *cos41° N = 2130.09 N
We apply Newton's first law for the balance in M1:
Σ Fy=0
Fn-W₁y=0 , Fn: normal force
Fn=W₁y=2130.09N
Friction Force = Ff=μs *Fn = 0.41*2130.09 =873.34 N
Σ Fx=0
T- W₁x- Ff=0
T= 1851.66 + 873.34
T= 1851.66 + 873.34
T=2725 N
We apply Newton's first law for the balance in M2:
Σ Fy=0
T- W₂ =0
W₂ = T = 2725 N
W₂ = M2*g
M2 = W₂/g
M2 = 2725/9.8
M2 = 278.06 kg