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Gala2k [10]
3 years ago
5

A motor vehicle has a mass of 1200kg and the road wheels have a radius of 360mm. The engine rotating parts have a moment of iner

tia of 0.36kgm2 and the four wheels together have a moment of inertia of 4.8kgm2. The gear ratio from engine to back axle is 6 and when the vehicle speed is v m/s, the engine torque available for propulsion is 82Nm and the resistance to motion is (240 +0?)N. Determine the acceleration of the vehicle at a speed of 15m/s on a level road and the time required to increase the speed of the vehicle from 15 to 25m/s​
Engineering
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer:

manda a senha do brainly bloquearam os amigos e você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo uma academia de lima e você vai ver o meu perfil completo a minha amiga Jeciane estamos completo a minha amiga Jeciane estamos fazendo uma academia de lima e você vai ver o meu perfil você vai ver o meu perfil você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo

Explanation:

Marcar como melhor porfavo

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How do you determine when a truss will fail?<br> (Yield Stress)
Marat540 [252]

Answer:

If a truss buckles or overturns, it is usually because of the failure of an adjacent truss or its bracing. A steel truss in a fire may buckle and overturn because of expansion or weakening from the heat. Most truss failures are the result of broken connections. Photo 1 shows a set of parallel-chord wood trusses supporting a plywood floor deck.

Explanation:

7 0
2 years ago
Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A
uysha [10]

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C   ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

7 0
3 years ago
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi
Anon25 [30]

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

\sigma = \frac{3 F L}{2 bd^{2} }

\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }

\sigma = 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

4 0
2 years ago
Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin
Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

So, $Q_2= \frac{1000}{100}=10 \ mm^3/s$

Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$    ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$  ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

       $=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

       $=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

       $=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

7 0
3 years ago
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0
2 years ago
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