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2 years ago

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Physics

2 answers:

madam [21]2 years ago

4 0

Answer:

Calculate the acceleration if it took 2 seconds to slow your car down.Explanation:

skelet666 [1.2K]2 years ago

4 0

I need points dog. I am sorry about the inconvenience michael.

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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

A woman pushes a 27 kg lawnmower at a steady speed. She exerts a 120 N force in a direction 35◦ below the horizontal. The accele

nikklg [1K]

The vertical force exerted on the lawn is 68.8 N downward

Explanation:

The vertical force exerted by the lawnmower on the lawn is equal to the vertical component of the force applied, therefore:

$F_y = F sin \theta$

where

F is the magnitude of the force applied

$\theta$ is the angle between the direction of the force and the horizontal

In this problem:

F = 120 N

$\theta=-35^{\circ}$

Substituting,

$F_y = (120)(sin 30)=-68.8 N$

where the negative sign means the direction of the force is downward.

Learn more about vector components and forces here:

brainly.com/question/2678571

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

3 0

3 years ago

Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un

Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

$K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}$

$\gamma$ = 1, $K_{IC}$ = 24.1

[/tex]\sigma_{y}[/tex] = 570

and, $\sigma_{f} = 570 \times \frac{3}{4}$

= 427.5

Hence, we will calculate the critical crack length as follows.

a = $\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}$

= $\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}$

= $10.13 \times 10^{-4}$

Therefore, largest size is as follows.

Largest size = 2a

= $2 \times 10.13 \times 10^{-4}$

= $20.26 \times 10^{-4}$

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is $20.26 \times 10^{-4}$ .

4 0

3 years ago

Problem 3.

Korolek [52]

Answer:

143.352 watt.

Explanation:

So, in the question above we are given the following parameters or data or information that is going to assist us in answering the question above efficiently. The parameters are:

"A 1.8 m wide by 1.0 m tall by 0.65m deep home freezer is insulated with 5.0cm thick Styrofoam insulation"

The inside temperature of the freezer = -20°C.

Thickness = 5.0cm = 5.0 × 10^-2 m.

Step one: Calculate the surface area of the freezer. That can be done by using the formula below:

Area = 2[ ( Length × breadth) + (breadth × height) + (length × height) ].

Area = 2[ (1.8 × 0.65) + (0.65 × 1.0) + (1.8 × 1.0)].

Area = 7.24 m^2.

Step two: Calculate the rate of heat transfer by using the formula below;

Rate of heat transfer =[ thermal conductivity × Area (T1 - T2) ]/ thickness.

Rate of heat transfer = 0.022 × 7.24(25+20)/5.0 × 10^-2 = 143.352 watt.

8 0

3 years ago

What is stopping distance of a body at 20 m/s is decelerated at 2m/s² to rest?<br>

AlladinOne [14]

Answer:

s = 100 m

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s = (0² - 20²) / 2(-2)

s = 100 m

5 0

3 years ago