Answer:
c. length of the wall or column and the rate of placement of the concrete
Explanation:
when designing for wall and column form-works, it is of utmost important to know the length of the wall and the type of concrete placement to be used.
Concrete placement has methods and precaution to be taken when doing the form work
if the concrete placement is manually (hand or funnel) the form work height should not be more than 1 m to enable easy compaction and vibration of concrete in the form.
Also, if the form work length is too long and it is not well reinforced, it tends to burg if the force apply during concrete placement or during vibration is much.
<h3><u>Answer;</u></h3>
<em>Electrons </em>
<h3><u>Explanation;</u></h3>
- <em><u>Thomson contributed to the model of an atom by discovery of </u></em><em><u>electrons </u></em><em><u>and thus proving the existence of sub-atomic particles in an atom. </u></em>
- <u><em>Thomson used cathode ray tube, and demonstrated that cathode rays were negatively charged.</em></u> According to his model normally known as the plum pudding in which he stated that an atom is composed of electrons as subatomic particles that are surrounded by positive charges to balance the electrons.
To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is
Here
Mass inside the orbit in terms of Volume and Density is
Where,
V = Volume
Density
Now considering the volume of the star as a Sphere we have
Replacing at the previous equation we have,
Now replacing the mass at the gravitational acceleration formula we have that
For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is
At the same time the general expression for the centripetal acceleration is
Where 
is the orbital velocity
Using this expression in the left hand side of the equation we have that
Considering the constant values we have that
As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.
So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density
Answer:
5.51 m/s^2
Explanation:
Initial scale reading = 50 kg
assume the greatest scale reading = 78.09 kg
<u>Determine the maximum acceleration for these elevators</u>
At rest the weight is = 50 kg
Weight ( F ) = mg = 50 * 9.81 = 490.5 N<u>
</u>
<u>
</u>At the 10th floor weight = 78.09 kg
Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N
F = change in weight
Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)
50 * a = 275.61
Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2
Spring potential energy:
E = 0.5 * k * x²
k spring constant
x spring compression
x = √(2 * E / k) = 0.7
