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____ [38]
2 years ago
9

Reduce the force F ij = + (2 5 ) kN to point A(2m,3m) that acts on point B( 3m,5m) - .

Engineering
1 answer:
Alexxx [7]2 years ago
6 0

Given :

Force, \vec{F}= (2\hat{i} + 5\hat{j})\ kN.

Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )

To Find :

The work done by force F .

Solution :

Displacement vector between point A and B is :

\vec{d} = (3-2)\hat{i} + (5-3)\hat{j}\\\\\vec{d} = \hat{i} + 2\hat{j}

Now, we know work done is given by :

W = \vec{F}.\vec{d}\\\\W= (2\hat{i} + 5\hat{j}).(\hat{i}+\hat{2j})\\\\W = (2\times 1) +( 5\times 2) \ kJ\\\\W = 12 \ kJ

W = 12000 J

Therefore, work done by force is 12000 J .

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If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

\sigma_b =  1.37 psi

Bending stress at point 3.96 is \sigma_b = 1.37 psi

3 0
3 years ago
you have a permanent magnet whit remanent induction. than you cut a piec of that magnet. if you put that piece back into permane
ELEN [110]
Yes, it will. The induction will be the same as long as it’s put back together.
5 0
3 years ago
What is 7-?=4 i need help
aleksandr82 [10.1K]

Answer:

3

Explanation:

7-3=4

6 0
3 years ago
Read 2 more answers
What type of siege engines were used by Saladin to capture Jerusalem in 1187?
mariarad [96]

Answer:

LOTS

Explanation:

Catapults, Towers, and Trebuchets were all used by Saladin to capture Jerusalem in 1187

8 0
2 years ago
Use phasor techniques to determine the impedance seen by the source given that R = 4 Ω, C = 12 μF, L = 6 mH and ω = 2000 rad/sec
Zielflug [23.3K]

Answer:

Z = 29.938Ω ∠22.04°

I = 2.494A

Explanation:

Impedance Z is defined as the total opposition to the flow of current in an AC circuit. In an R-L-C AC circuit, Impedance is expressed as shown:

Z² = R²+(Xl-Xc)²

Z = √R²+(Xl-Xc)²

R is the resistance = 4Ω

Xl is the inductive reactance = ωL

Xc is the capacitive reactance =

1/ωc

Given C = 12 μF, L = 6 mH and ω = 2000 rad/sec

Xl = 2000×6×10^-3

Xl = 12Ω

Xc = 1/2000×12×10^-6

Xc = 1/24000×10^-6

Xc = 1/0.024

Xc = 41.67Ω

Z = √4²+(12-41.67)²

Z = √16+880.31

Z = √896.31

Z = 29.938Ω (to 3dp)

θ = tan^-1(Xl-Xc)/R

θ = tan^-1(12-41.67)/12

θ = tan^-1(-29.67)/12

θ = tan^-1 -2.47

θ = -67.96°

θ = 90-67.96

θ = 22.04° (to 2dp)

To determine the current, we will use the relationship

V = IZ

I =V/Z

Given V = 12V

I = 29.93/12

I = 2.494A (3dp)

7 0
3 years ago
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