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4 years ago

A automotive test driver travels due north in a prototype hybrid vehicle at 30 mi/h

Physics

1 answer:

Nikolay [14]4 years ago

3 0

Answer:

Option A. 40 mi/h

Explanation:

To obtain the average speed of the vehicle, we'll begin by calculating the distance travelled by the vehicle in each case. This is illustrated below:

Case 1:

Speed = 30 mi/h

Time = 2 h

Distance =...?

Speed = Distance /Time

30 = Distance /2

Cross multiply

Distance = 30 × 2

Distance = 60 mi

Case 2:

Speed = 60 mi/h

Time = 1 h

Distance =...?

Speed = Distance /Time

60 = Distance /1

Cross multiply

Distance = 60 × 1

Distance = 60 mi

Finally, we shall determine the average speed of the vehicle as follow:

Total distance travelled = 60 + 60

Total distance travelled = 120 mi

Total time = 2 + 1

Total time = 3 h

Average speed =..?

Average speed = Total Distance travelled /Total time

Average speed = 120/3

Average speed = 40 mi/h

Therefore, the average speed of the vehicle is 40 mi/h

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Sav [38]

Answer:

65 Newtons

Explanation:

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3 years ago

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Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

WILL GIVE BRAINLIEST

Diano4ka-milaya [45]

Answer:

nice to help you

Explanation:

If for any reason the Sun shrank smaller than the Earth, this shrunken Sun wouldn't have the mass to create fusion and would burn out completely. Our solar system would lose its only star. ... Earth's mass would be at least 333,000 times bigger than it is now. Imagine the gravity that planet would have.

5 0

3 years ago

A sealed pressure cooker allows steam to escape only through an opening in the lid. a separate metal object, called a petcock, s

daser333 [38]

The mass of the petcock of the pressure cooker is 0.257 kg.

The total pressure inside the cooker is= atmospheric pressure+Gauge pressure

=100 kPa+101 kPa=201*10^3 Pa

The area of the petcock is= (π/4)d^2=0.785*0.004^2=12.56*10^(-6) m^2

Now the mass of the petcock is calculated as

P=(F/A)=(mg/A)

201*10^3=(m*9.81/(12.56*10^(-6)

m=0.257 kg

Therefore the mass of the petcock is 0.257 kg

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4 years ago

Effect of thunder storm

tangare [24]

Answer:

Some of the most severe weather occurs when a single thunderstorm affects one location for an extended time. Thunderstorms can bring heavy rains (which can cause flash flooding), strong winds, hail, lightning, and tornadoes. Severe thunderstorms can cause extensive damage to homes and property.

Explanation:

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3 years ago