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3 years ago

The Texas Sure program is designed to:

1 answer:

4 0

TexasSure is designed to reduce the number of uninsured drivers and cut costs for responsible Texans, who now pay almost $900 million a year to protect themselves against those with no coverage. Currently, an estimated 20 percent of Texas drivers are uninsured.

TexasSure is the financial responsibility verification program created by the 79th Texas Legislature, Regular Session, in Senate Bill 1670.

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Which of the following has special properties that allow forces and pressure to be distributed evenly?

Thepotemich [5.8K]

Answer:

Fluids

Explanation:

Fluids has special properties that allow forces and pressure to be distributed evenly within them.

Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
Therefore, when pressure is applied to them, it permeates evenly on all parts.
Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.

7 0

3 years ago

In order to impress your neighbors and improve your vision in traffic jams, you decide to mount a cylindrical periscope 2.0 m hi

kondaur [170]

Follow @richard.gbe on Instagram for the answer

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4 years ago

A wall that cannot be moved because it is carrying the weight of the roof is considered a wall

ad-work [718]

Answer:

Blank wall

Explanation:

A wall that cannot be moved because it is carrying the weight of the roof is considered a blank wall.

5 0

2 years ago

Air at 80 kPa and 10Â°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre

il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

$A_2=5A_1$

If consider that density of air is not changing from inlet to exit then by using continuity equation

$A_1V_1=A_2V_2$

So $A_1\times 150=5A_1V_2$

$V_2=30$ m/s

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

Here Q=0 and w=0

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

When air is treating as ideal gas

$h=C_pT$

Noe by putting the values

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

$1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}$

$T_2=293.74K$

So the exit temperature is 293.74 K.

7 0

3 years ago

A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 Mpa root m is exposed to a stress of 1030 MPa. W

cupoosta [38]

Answer:

It will not experience fracture when it is exposed to a stress of 1030 MPa.

Explanation:

Given

Klc = 54.8 MPa √m

a = 0.5 mm = 0.5*10⁻³m

Y = 1.0

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of KIc, Y, and the largest value of a in the material. This requires that we solve for σc from the following equation:

σc = KIc / (Y*√(π*a))

Thus

σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))

⇒ σc = 1382.67 MPa > 1030 MPa

Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.

3 0

3 years ago