10.16.1: LAB: Interstate highway numbers (Python)

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of t

Tju [1.3M]

Answer:

15.99 ft/s

Explanation:

From Newton's equation of motion, we have

v = u + at

v = Final speed

u = initial speed

a = acceleration

t = time

now

for the points A and C

v = 17.6 ft/s

u = 13.2 ft/s

t = 3 s

thus,

17.6 = 13.2 + a(3)

or

3a = 17.6 - 13.2

3a = 4.4

or

a = 1.467 m/s²

Thus,

For Points A and B

v = speed at B i.e v'

u = 13.2 ft/s

a = 1.467 ft/s²

t = 1.90 s

therefore,

v' = 13.2 + (1.467 × 1.90 )

v' = 13.2 + 2.7867

v' = 15.9867 ≈ 15.99 ft/s

3 0

3 years ago

If a plus sight of 12.03 ft is taken on BM A, elevation 312.547 ft, and a minus sight of 5.43 ft is read on point X, calculate t

DochEvi [55]

Answer:

Therefore, height of instrument is 324.577 ft

Therefore, elevation of point x is 330 m

Explanation:

Given that

Plus sight on BM = 12.03 ft

Minus sight is = 5.43 ft

Elevation = 312.547 ft

Height of instrument is H.I

H.I = elevation on bench mark + plus sight

= 312.547 + 12.03 = 324.577 ft

Therefore, height of instrument is 324.577 ft

Elevation at point x is = H.I - minus sight

= 324.577 - (- 5.43)

= 330.00 m

Therefore, elevation of point x is 330 m

3 0

4 years ago

Are considered a form of personal protective equipment (PPE) for eyes.

vampirchik [111]

Answer:

All of the above

Explanation:

6 0

3 years ago

If a condenser has high head pressure and a higher than normal temperature, a technician could ____.

katrin2010 [14]

clean the tubes and fins with a high-pressure jet of air or mechanical scrubbing

ensure that the condenser fans are operating properly

6 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

4 years ago