Answer:
a)
, b) 
Explanation:
a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:




b) The speed of the block is determined by using the Principle of Energy Conservation:




The radius of the circular loop is:



Answer:
I think it's D) Guess why? I searched it up!
Answer:
F_net = 26.512 N
Explanation:
Given:
Q_a = 3.06 * 10^(-4 ) C
Q_b = -5.7 * 10^(-4 ) C
Q_c = 1.08 * 10^(-4 ) C
R_ac = 3 m
R_bc = sqrt (3^2 + 4^2) = 5m
k = 8.99 * 10^9
Coulomb's Law:
F_i = k * Q_i * Q_j / R_ij^2
Compute F_ac and F_bc :
F_ac = k * Q_a * Q_c / R^2_ac
F_ac = 8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2
F_ac = 33.01128 N
F_bc = k * Q_b * Q_c / R^2_bc
F_bc = 8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2
F_bc = - 22.137 N
Angle a is subtended between F_bc and y axis @ C
cos(a) = 3 / 5
sin (a) = 4 / 5
Compute F_net:
F_net = sqrt (F_x ^2 + F_y ^2)
F_x = sum of forces in x direction:
F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N
F_y = sum of forces in y direction:
F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N
F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N
Answer: F_net = 26.512 N
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