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aev [14]
4 years ago
12

The cheetah is one of the fastest-accelerating animals, because it can go from rest to 19.6 m/s (about 44 mi/h) in 2.9 s. If its

mass is 108 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
(a) watts(b) horsepower.
Physics
1 answer:
Vitek1552 [10]4 years ago
5 0

Answer:

a)P =14288.4 W

b)P = 19.16  horsepower

Explanation:

Given that

m= 108 kg

Initial velocity ,u= 0 m/s

Final velocity ,v= 19.6 m/s

t= 2.9 s

Lets take acceleration of Cheetah is a m/s²

We know that

v= u  + a t

19.6 = 0 + a x 2.9

a= 6.75 m/s²

Now force F

F= m a

F= 108 x 6.75 N

F= 729 N

Now the power P

P = F.v

P = 729 x 19.6 W

P =14288.4 W

We know that

1 W= 0.0013  horsepower

P = 19.16  horsepower

P =14288.4 W

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Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They
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Answer:

Explanation:

a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s.

F = M x 48

Mass M = F / 48

a )

When force = 2F and mass = M

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= 2F /F/48

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b )

When force = F and mass = 2M

Acceleration = force / mass

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=  24 cm/s²

c )

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Acceleration = force / mass

= 2F /2F/48

=  48  cm/s

d )

When force = 2F and mass = 4M

Acceleration = force / mass

= 2F /4F/48

=  24  cm/s

e)

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7 0
4 years ago
Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.4 m/s. (a) Determine the appr
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Answer:

1.12 m

0.08291 m

Explanation:

u = Upstream velocity = 0.4 m/s

Re = Reynold's number = 5\times 10^5 (turbulent)

\nu = Viscosity of water = 1.12\times 10^{-6}\ Pas

Here the flow is turbulent so we have the relation

Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m

Boundary layer thickness relation is given by

\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m

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What is the name given to group 18 on the periodic table?​
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