Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
<em><u>Support Cy:</u></em>
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
<em><u>Support Ay:</u></em>
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
a) The ball goes one-third times higher on X
b) The ball goes three times higher on X.
Explanation:
a)
- As the initial velocity is the same than on Earth, but the free-fall acceleration is three times larger, this means that the only net force acting on the ball (gravity) will be three times larger, so it is clear that the ball will reach to a lower height, as it will slowed down more quickly.
- Kinematically, as we know that the speed becomes zero when the ball reaches to the maximum height, we can use the following kinematic equation:
![v_{f} ^{2} - v_{o}^{2} = 2* \Delta h* g](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20-%20v_%7Bo%7D%5E%7B2%7D%20%20%3D%202%2A%20%5CDelta%20h%2A%20g)
since vf = 0, solving for Δh, we have:
![\Delta h = h_{max} =\frac{v_{o} ^{2}}{2*g} (1)](https://tex.z-dn.net/?f=%5CDelta%20h%20%3D%20h_%7Bmax%7D%20%3D%5Cfrac%7Bv_%7Bo%7D%20%5E%7B2%7D%7D%7B2%2Ag%7D%20%281%29)
if v₀ₓ = v₀E, and gₓ = 3*gE, replacing in (1), we get:
Δhₓ = 1/3 * ΔhE
which confirms our intuitive reasoning.
b)
- Now, if the initial velocity is three times larger than the one on Earth, even the acceleration due to gravity is three times larger, we conclude that the ball will go higher than on Earth.
- We can use the same kinematic equation as in (1) replacing Vox by 3*VoE, as follows:
![\Delta h = h_{max} =\frac{(3*v_{o}) ^{2}}{2*3*g} (2)](https://tex.z-dn.net/?f=%5CDelta%20h%20%3D%20h_%7Bmax%7D%20%3D%5Cfrac%7B%283%2Av_%7Bo%7D%29%20%5E%7B2%7D%7D%7B2%2A3%2Ag%7D%20%282%29)
Replacing the right side of (1) in (2), we get:
Δhx = 3* ΔhE
which confirms our intuitive reasoning also.
Answer:
This statement is False.
Explanation:
This is because the molecules of gases are randomly traveling and they even sometimes collide with each other. So, it is true that they resist crowding but it is not always possible for the gas molecules to be as far apart as possible because of their random movements. This also makes it impossible for the gas molecules in the tank to be always at the maximum possible distance from its nearest neighbor.
However, the distance between the electrons is not due to random movements like gases. Electrons are placed at the distance from each other due to the forces of repulsion between them, as the same charged particles always repel each other. So, the dispersion of the electron on the copper ball is not due to the random distribution of electrons. However, they will be placed at a distance to each other due to the force of repulsion between them.