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Dmitriy789 [7]
3 years ago
12

A race cars velocity increases from 6.0 m/s to 36 m/s over a 5.0 s time interval. What is It’s average acceleration?

Physics
1 answer:
Sonja [21]3 years ago
3 0

The acceleration of the race car is 6 m/s²

The given parameters;

initial velocity of the race car, u = 6 m/s

final velocity of the race car, v = 36 m/s

time of motion, t = 5 s

Acceleration is defined as the change in velocity per change in time of motion.

The acceleration of the race car is calculated as follows;

a=\frac{v-u}{t} \\\\a = \frac{36-6}{5} \\\\a = 6 \ m/s^2

Thus, the acceleration of the race car is 6 m/s²

Learn more here: brainly.com/question/17280180

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Answer:

The work done in winding the spring gets stored in the wound up spring in the form of elastic potential energy (i.e potential energy due to change in shape). ... During this process, the potential energy stored in it gets converted to kinetic energy. This turns the wheels of the toy car.

Explanation:

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During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc
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V o = 6 m/s,
t = 2 s
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3 years ago
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
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