Answer:
w = 11.211 KN/m
Explanation:
Given:
diameter, d = 50 mm
F.S = 2
L = 3
Due to symmetry, we have:



To find the maximum intensity, w, let's take the Pcr formula, we have:

Let's take k = 1
Substituting figures, we have:

Solving for w, we have:

w = 11211.14 N/m = 11.211 KN/m
Since Area, A= pi * (0.05)²
. This means it is safe
The maximum intensity w = 11.211KN/m
Answer:
Convection currents are the result of differential heating. Lighter (less dense), warm material rises while heavier (more dense) cool material sinks. It is this movement that creates circulation patterns known as convection currents in the atmosphere, in water, and in the mantle of Earth.
Explanation:
Hope this helps
Answer:
16.53 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 18.0 m/s.
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
The maximum height reached by the ball can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 18² – (2 × 9.8 × h)
0 = 324 – 19.6h
Rearrange
19.6h = 324
Divide both side by 19.6
h = 324 / 19.6
h = 16.53 m
Therefore, the maximum height reached by the ball is 16.53 m
Answer:
Explanation:
Force on the electron = q ( v x B )
q = - 1.6 x 10⁻¹⁹
v = (5.9i−6.4j)×10⁴
B = (−0.63i+0.65j)
v x B = (5.9i−6.4j)×10⁴ x (−0.63i+0.65j)
= (3.835 - 4.032 ) x 10⁴ k
= - 1970 k
Force on the electron = q ( v x B )
= - 1.6 x 10⁻¹⁹ x -1970 k
= 3.152 x 10⁻¹⁶ k
z-component of the force on the electron
Fz = 3.152 x 10⁻¹⁶ N
Answer:
Explanation:
Velocity of Top most point of wheel is twice the Velocity of centre of mass of wheel

Thus angular velocity is given by



![K.E.=\frac{1}{2}m_{stone}V^2+2\left [ \frac{1}{2}m_{cyl}\left [ \frac{V}{2}\right ]^2\right ]+2\left [ \frac{1}{2}I\omega ^2\right ]](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B1%7D%7B2%7Dm_%7Bstone%7DV%5E2%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7Dm_%7Bcyl%7D%5Cleft%20%5B%20%5Cfrac%7BV%7D%7B2%7D%5Cright%20%5D%5E2%5Cright%20%5D%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%20%5E2%5Cright%20%5D)
![K.E.=\frac{V^2}{2}\left [ m_{stone}+\frac{m_{roller}}{2}\right ]+I\omega ^2](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7BV%5E2%7D%7B2%7D%5Cleft%20%5B%20m_%7Bstone%7D%2B%5Cfrac%7Bm_%7Broller%7D%7D%7B2%7D%5Cright%20%5D%2BI%5Comega%20%5E2)
![K.E.=\frac{0.319^2}{2}\left [ 672+\frac{82}{2}\right ]+\frac{82\times 0.343^2\times 0.465^2}{2}](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B0.319%5E2%7D%7B2%7D%5Cleft%20%5B%20672%2B%5Cfrac%7B82%7D%7B2%7D%5Cright%20%5D%2B%5Cfrac%7B82%5Ctimes%200.343%5E2%5Ctimes%200.465%5E2%7D%7B2%7D)
