Answer:F=400.28 N
Explanation:
Given
mass of large cube 
mass of small cube 
coefficient of static friction 
acceleration a of the system is given by


Now Normal reaction between two blocks is given by

friction tries to balance weight
thus 
and 






Answer:
a) v = (-7.73 i ^ - 6.95j ^) 10⁶ m/s and b) ΔK = 3.96 10⁻¹³ J
Explanation:
We can work this process of disintegration as a conservation problem of the moment.
We create a system formed by the initial nucleus and the three final particles, in this case all the forces involved are internal and the amount of movement is conserved. We will write the equations each axis
X axis
p₀ = 0
pf = m1 v1x + m vfx
m₁ = 8.50 10-27 kg
vf₁ = 4.00 106 m / s
p₀ = pf
0 = m₁ v₁ₓ + m₃ vfₓ
vfₓ = -m₁ / m₃ v1ₓ
Y Axis
P₀ = 0
Pf = m₂ v₂ + m₃ vfy
m₂ = 5.10 10⁻²⁷ kg
v₂ = 6.00 10⁶ m / s
p₀ = pf
0 = m₂ v₂ + m₃ vfy
vfy = -m₂ / m3₃v₂
We have the initial particle mass and it decomposes into three parts after disintegration
m = m₁ + m₂ + m₃
m₃ = m-m₁-m₂
m₃ = 1.80 10⁻²⁶ - 5.10 10⁻²⁷ - 8.50 10⁻²⁷ = (1.80 -0.510 -0.850) 10⁻²⁶
m3 = 0.44 10⁻²⁶ kg = 4.4 10⁻²⁷ kg
Let's replace and calculate
vfₓ = -m₁ / m₃ v₁ₓ
vfₓ = - 8.50 10⁻²⁷/4.4 10⁻²⁷ 4.00 10⁶
vfₓ = -7.73 10⁶ m / s
vfy = -m₂ / m₃ v₂
vfy = - 5.10 10⁻²⁷ /4.4 10⁻²⁷ 6.00 10⁶
vfy = -6.95 10⁶ m/s
We set the speed vector
v = (-7.73 i ^ - 6.95j ^) 10⁶ m/s
b) Let's calculate the kinetic energy
Initial
K₀ = 0
Final
Kf = K₁ + K₂ + K₃
Kf = ½ m₁ v₁² + ½ m₂ v₂² + ½ m₃ v₃²
v₃² = (7.73 10⁶)²+ (6.95 10⁶)² = 108.5 10⁻¹²
Kf = ½ 8.50 10⁻²⁷(4 10⁶) 2 + ½ 5.1 10⁻²⁷ (6 10⁶) 2 + ½ 4.4 10⁻²⁷ 108.05 10⁻¹²
Kf = 68 10⁻¹⁵ + 91.8 10⁻¹⁵ + 237.7 10⁻¹⁵
Kf = 396.8 10⁻¹⁵ J
Kf = 3.96 10⁻¹³ J
ΔK = Kf -K₀
ΔK = 3.96 10⁻¹³ J
Answer:
No one is right
Explanation:
John Case:
The function
is defined between -1 and 1, So it is not possible obtain a value
greater.
In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.
Larry case:
Is you have
, the domain of this is [0,2].
it is equivalent to adding 1 to the domain of the
, and its mean that the function
, in general, is not greater than
.
The two factors are mass and distance between them.
Given :
Mass of person, m = 81 kg.
Area of high heal, A = 3.5 cm² = 0.00035 m².
To Find :
The pressure applied by the heal on the ground.
Solution :
We know, pressure is given by :

Hence, this is the required solution.