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3 years ago

0/5 pts

Engineering

1 answer:

Brilliant_brown [7]3 years ago

4 0

Explanation:

150 divide by 150 and that how you do the is you what to divide together 15/ 150 you welcome have a good day is you need something else

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A particle travels along a straight line with a velocity v = (12 – 3t2) m/s. When t = 1 s, the particle is located 10 m to the l

arlik [135]

Answer:

The displacement from t = 0 to t = 10 s, is -880 m

Distance is 912 m

Explanation:

$v = (12 - 3t^2) m/s = ds/dt$ . . . . . . . . . . A

integrate above equation we get

$s = 12t - t^3 + C$

from information given in the question we have

t = 1 s, s = -10 m

so distance s will be

-10 = 12 - 1 + C,

C = -21

$s(t) = 12t - t^3 - 21$

we know that acceleration is given as

$a(t) = dv/dt = -6t$

[FROM EQUATION A]

Acceleration at t = 4 s, a(4) = -24 m/s^2

for the displacement from t = 0 to t = 10 s,

$s(10) - s(0) = (12*10 - 10^3 - 21) - (-21) = -880 m$

the distance the particle travels during this time period:

let v = 0,

$3t^2 = 12$

t = 2 s

Distance $= [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m$

7 0

4 years ago

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part

ra1l [238]

Answer:

h1 = 290.16kj/kg

P = 1.2311

Prandil expression at 8

P=p1/p7×pr

=8(1.2311)

=9.85

Enthalpy state at 8 corresponding to 9.85

h1 = 526.13kj/kg

Now prandtl state at 9 that correspond to 1400k.

h9 = 1515.42kj/kg

Pr = 450.5

Prandtl expression at state 10

P= p10/p9×pr

=1/8(450.5)

=56.31

Enthalpy at state 10 corresponding to prandtl 56.31

h10 = 860.39kj/kg

At 520k

h11 = 523.63kj/kg

4 0

3 years ago

A water tank is completely filled with liquid waterat 20°C.The tank material is such that it can withstand tensioncaused by a vo

Xelga [282]

Answer:

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Explanation:

From literature, the coefficient of volume expansion of water between 20°C and 50°C = β = (0.377 × 10⁻³) K⁻¹

Volume expansivity is given by

ΔV = V β ΔT

ΔV = Change in volume

V = initial volume

β = Coefficient of volume expansion = (0.377 × 10⁻³) K⁻¹ = 0.000377 K⁻¹

ΔT = Change in temperature = ?

It is given in the question that maximum volume increase the tank can withstand is

(ΔV/V) × 100% = 0.8%

(ΔV/V) = 0.008

V β ΔT = ΔV

β ΔT = (ΔV/V)

β ΔT = 0.008

ΔT = (0.008/β)

ΔT = (0.008/0.000377)

ΔT = 21.22°C

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Hope this Helps

5 0

4 years ago

I need help with my autos

oksano4ka [1.4K]

Answer:

what is wrong with it and what is the question

Explanation:

6 0

3 years ago

A rigid canister with a radius of 5 in and a height of 10 in is filled with air. The initial pressure and temperature of air in

Elza [17]

Answer:1.458 Btu

Explanation:

Given

radius of canister $\left ( r\right )=5in$

Height of canister $\left ( h\right )=10 in.$

Initial pressure $\left ( P_i\right )=14.7 Psi$

Initia ltemprature $\left ( T_i\right )=70^{\circ}F$

Final pressure $\left ( P_f\right )=30Psi$

as canister is rigid therefore change in volume is zero

therefore

$\frac{P_i}{T_i}$ = $\frac{P_f}{T_f}$

$\frac{14.7}{70}$ = $\frac{30}{T_f}$

$T_f=142.85^{\circ}F$

volume of canister= $\pi \times r^{2}\times h$

= $\pi \times 5^{2}\times 10=250\pi$

volume of canister=12872,038.8 $mm^3$

now calculating mass of air

PV=mRT

substituting values

$\left ( 14.7Psi\right )\left ( 12872,038.8 mm^3\right )=m\left ( 0.287\right )\left ( 70^{\circ}F\right )$

m=21.0192 gm

Therefore heat transferred = $mc_p\left ( T_f-T_i\right )$

= $21.0192\times 10^{-3}\times \left ( 142.857-70\right )$

=1539.052J=1.458Btu

3 0

4 years ago