Complete Question
A force of F =460 lb. acts on the curved member as shown. Determine the resultant normal force across the cross section at point B. Express your answer to three significant figures and include appropriate units.(Figures Attached)
Answer:
Resultant Normal Force = 55. 02 lb.
Explanation:
In Order to find the resultant normal force we must first resolve the figure to find out the x and y components of the force across the cross-section at point B. Hence after resolving the figure we can find the resultant normal force by finding the vertical component of the force: ∑ F_y
The normal force acting at point B will be:
∑ F_y :-
N_F + 460(4/5)sin〖30〗^° - 460(3/5) sin〖60〗^° = 0
N_F + 184 - 239.02 = 0
N_F = 55. 02 lb.
Solution :
The nuclear reaction for boron is given as :
And the reaction for Cadmium is :
We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.
Answer:
<em>number of mole is 31342.36 moles</em>
<em>mass is 125.369 kg</em>
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Explanation:
Diameter of the spherical balloon d = 9 m
radius r = d/2 = 9/2 = 4.5 m
The volume pf the sphere balloon ca be calculated from
V =
V = = 381.75 m^3
Temperature of the gas T = 20 °C = 20 + 273 = 293 K
Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa
number of moles n = ?
Using
PV = nRT
where
P is the pressure of the gas
V is the volume of the gas
n is the mole number of the gas
R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1
T is the temperature of the gas (must be converted to kelvin K)
substituting values, we have
200 x 10^3 x 381.75 = n x 8.314 x 293
number of moles n = 76350000/2436 = <em>31342.36 moles</em>
We recall that n = m/MM
or m = n x MM
where
n is the number of moles
m is the mass of the gas
MM is the molar mass of the gas
For helium, the molar mass = 4 g/mol
substituting values, we have
m = 31342.36 x 4
m = 125369.44 g
m =<em> 125.369 kg</em>
Answer:
The total hole mobility is 41.67 cm²/V s
Explanation:
Data given by the exercise:
hole mobility due to lattice scattering = ul = 50 cm²/V s
hole mobility due to ionized impurity = ui = 250 cm²/V s
The total mobility is equal: