The catalytic reaction base of ester and amide hydrolysis. the corrosive acids will then react with the skin and hydrolyze the fats which are chemical forms of esters.
Answer:
The speed with which just after he grabs her is 2.68 m/s.
Explanation:
Given that,
Mass of Erica, m = 38 m/s
Mass of Danny, m' = 46 kg
Erica reaches the high point of her bounce, Danny is moving upward past her at 4.9 m/s. At this moment, the initial speed of Erica will be 0. The momentum will remain conserved. Using the conservation of linear momentum as :

So, the speed with which just after he grabs her is 2.68 m/s. Hence, this is the required solution.
A rotation motion is a motion that takes place around a fixed axis.
Like gears turning on each other.
Answer:
Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]
Explanation:
To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.
The gravitational force is equal to:
Fg = (10 * 9.81) = 98.1 [N]
Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.
N - Fg = 0
N = Fg
N = 98.1 [N]
The friction force is defined as the product of normal force by the coefficient of friction.
Ff = N * μ
Ff = 98.1 * 0.4
Ff = 39.24 [N]
By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.
60 - Ff = m * a
60 - 39.24 = 10 * a
a = 2.076[m/^2]
Answer:
Explanation:
Given that,
Current in loops are
i1 = 12A
i2 = 20A
The loops are 3.4cm apart
The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop
Magnetic Field is given as
B= μoi/2πr
Where,
μo is a constant = 4π×10^-7 Tm/A
r is the distance between the two wires
i is the current in the wires
B is the magnetic field
NOTE
Field due to large loop should be equal to the smaller loop.
B1 = B2
μo•i1 / 2π•r1 = μo•i2 / 2π•r2
Then, μo, 2π cancels out, so we have
i1 / r1 = i2 / r2
Make r2 subject of formula
i1•r2 = i2•r1
r2 = i2•r1 / i2
r2 = 20×3.4/12
r2 = 5.67cm
The radius of the bigger loop is 5.67cm.