Occupational
Good luck I just copied the guy I front lol
Answer:
The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons
Explanation:
We know that for two point charges of magnitude 
the magnitude of force between them is given by
where
is constant
is the separation between the charges
Initially when the charges are separated by 2.4 meters the force can be calculated as
Now when the separation is reduced to 0.7 meters the force is similarly calculated as
Applying value of the constant we get
Thus 
Answer:
1 The weight of the foundation block should be enough to withstand vibrations
2 The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.
3.The base should be sufficiently rigid to have the requisite rigidity
4 a distance should be created all across machine foundation to separate it from the adjacent parts of the building
Explanation:
1. The weight of the foundation block should be enough to withstand vibrations and to avoid friction between device and the surrounding soil as well. This can be done by increasing the base block weight in supporting with engine power
2.The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.
3.The base should be sufficiently rigid to have the requisite rigidity as the slightest misdirection of foundation may cause significant bearing disorders.
4.To avoid propagation of vibration from a device to the adjacent parts of the building, a distance should be created all across machine foundation to separate it from the adjacent parts of the building.
Answer:
P=3.31 hp (2.47 kW).
Explanation:
Solution
Curve A in Fig1. applies under the conditions of this problem.
S1 = Da / Dt ; S2 = E / Dt ; S3 = L / Da ; S4 = W / Da ; S5 = J / Dt and S6 = H / Dt
The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.
32.2
Fig. 32.2 Dimension of turbine agitator
The Reynolds number is calculated. The quantities for substitution are, in consistent units,
D a =2⋅ft
n= 90/ 60 =1.5 r/s
μ = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s
ρ = 93.5 lb/ft3 g= 32.17 ft/s2
NRc = (( D a) 2 n ρ)/ μ = 2 2 ×1.5×93.5 8.06× 10 −3 =69,600
From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P × (n) 3 × ( D a )5 × ρ g c
The power P= 5.8×93.5× (1.5) 3 × (2) 5 / 32.17 =1821⋅ft−lb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).
Answer:
(C) Prototype Model
Explanation:
I'm sure that is the answer i am very sorry if not :)
