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3 years ago

Is a jeep cherokee faster than a bmw 325i

Engineering

2 answers:

soldier1979 [14.2K]3 years ago

8 0

Answer:

Yes

Explanation:

IrinaK [193]3 years ago

7 0

Yes it is because of how fast it goes

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A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.

ludmilkaskok [199]

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

D=200 mm

P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

hoop stress $\sigma _{h}=\frac{Pd}{2t}$

Longitudinal stress $\sigma _{l}=\frac{Pd}{4t}$

So maximum shear tress in plane $\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}$

$\tau _{max}=\dfrac{Pd}{8t}$

Now by putting the value

$27.5=\dfrac{4\times 200}{8t}$

So t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

4 0

3 years ago

The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc

Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.

Given Data

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist

∅=4°

∅=4* $\pi$ /180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist

∅=TL/GJ

∅=TL/G* $\pi$ /2*c^4

∅=2TL/G* $\pi$ *c^4

c= $\sqrt[4]{2TL/\pi G }$ ∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J* $\pi$ /2*c^4)]*c

г =[2T/(J* $\pi$ *c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required

We will use larger of the two values

c= 18.0569 x 10^-3 m

c = 18.0569 mm

3 0

3 years ago

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

Dafna11 [192]

Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

4 0

3 years ago

Gaining unauthorized access to a computer's data is called (5 points)

yanalaym [24]

Hacking is correcttttttttt

6 0

3 years ago

Vehicles begin to arrive at a parking lot at 8:10 am at a constant rate of 6 veh/min until 8:25 am. There is no arrival from 8:2

tresset_1 [31]

1.i am superman

2.175 175 175

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4.bye classmate magbabalik pa ako

8 0

2 years ago