Answer:
Drawing the triangle:
H / x = tan 52.2 = 1.29
H / (4.6 - x) = tan 28.8 = .550
H = 1.29 x
H = .55 * 4.6 - .55 x
1.84 x = 2.53 combining equations
x = 1.38
4.6 - 1.38 = 3.22
Total base of triangle = 1.38 + 3.22 = 4.6
H / x = tan 52,2 = 1.29
H = 1.29 * 1.38 = 1.78 height of triangle
Check:
1.78 / 3.22 = tan 28.9
This agrees with the given value of 28.8
Answer:
We know there's two forces acting on a book while it sits on a table:the force of gravity pulling it down, and the normal force of the table acting upward on the book. The book isn't accelerating while it sits there. That's because the weight of the book is being counteracted by the normal force of the table.
Explanation:
There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.
By giving them an advice and by giving them encouraging and explaining the any theory
The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.
Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get
A. The proeutectoid phase is Fe₃c because 0.95 wt/c is greater than the eutectoid composition which is 0.76 wt/c
b. We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is
(0.14) × (3.5kg) = 0.49kg
c. We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.
