What does it mean when the service engine light is on?

Whats the best used for cable -stayed bridge

nalin [4]

Answer:

a cable -stayed bridge has, one or more towers,from which cable support the bridge deck.

7 0

3 years ago

A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

$F_f= \mu N_2$

where :

$F_f =$ friction force

$\mu$ = coefficient of friction

$N_2$ = normal friction

Given that:

$\mu$ = 0.3

$F_f =$ 0.3 $N_2$

Using the equation of equilibrium along horizontal direction.

$\sum f_x = 0$

P - $F_f =$ 0

P = 0.3 $N_2$ ----- Equation (1)

To determine the moment about point B ; we have the expression

$\sum M_B = 0$

0 = $N_2*70-W*35-P*100$

where;

P = horizontal force

$N_2$ = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = $N_2*70-650*35-100*P$

$P = \frac{70 N_2-22750}{100} ----- equation (2)$

Replacing $\frac{70 N_2-22750}{100}$ for P in equation (1)

$\frac{70N_2 -22750}{100} =0.3 N_2$

$N_2 = \frac{22750}{40}$ $N_2 = 568.75 \ N$

Plugging the value of $N_2 = 568.75 \ N$ in equation (2)

$P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}$

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b) From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

$\sum M_A = 0$

-(P × 100)+ (W×35) = 0

P = $\frac{W*35}{100}$

Replacing 650 N for W

$P = \frac{650*35}{100}$

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

5 0

3 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

An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t

11Alexandr11 [23.1K]

Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

$E=h/t = \frac{(1/2\pi)*6.626*10x^{-34} J.s}{20*10x^{-9} } = 5.273*10x^{-27} J = 3.29* 10^{-8} eV$ .

The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

λ = c/v

dλ/dv = -c/v² = -λ²/c

Therefore,

∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

= 9.20 × 10⁻¹⁵ m or 9.20 fm

The length of the photon (<em>l)</em> is

l = (light velocity) × (emission duration)

= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m

6 0

3 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

artcher [175]

Answer:

MUDA MUDA A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

4 0

3 years ago