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Anna71 [15]
3 years ago
15

a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin

al KE?(unit=J)PLEASE HELP
Physics
1 answer:
Sveta_85 [38]3 years ago
6 0

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

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A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate
yaroslaw [1]

Answer:

The  maximum error is  \Delta  \eta  = 2032.9

Explanation:

From the question we are told that

     The length  is  l  =  1\ m

      The radius is  r =  0.002 \pm  0.0002 \ m

        The pressure is  P  =  4 *10^{5} \ \pm 1750

        The  rate  is  v =  0.5*10^{-9} \ m^3 /t

       The viscosity is  \eta  =  \frac{\pi}{8} * \frac{P *  r^4}{v}

The error in the viscosity is mathematically represented  as

       \Delta  \eta  = | \frac{\delta \eta}{\delta P}| *  \Delta  P   +    |\frac{\delta \eta}{\delta r} |*  \Delta  r +  |\frac{\delta \eta}{\delta v} |*  \Delta  v

   Where  \frac{\delta \eta }{\delta P} =  \frac{\pi}{8} *  \frac{r^4}{v}

and         \frac{\delta \eta }{\delta r} =  \frac{\pi}{8} *  \frac{4* Pr^3}{v}

and          \frac{\delta \eta }{\delta v} =  - \frac{\pi}{8} *  \frac{Pr^4}{v^2}

So  

             \Delta  \eta  = \frac{\pi}{8} [ |\frac{r^4}{v}  | *  \Delta  P   +    | \frac{4 *  P * r^3}{v}  |*  \Delta  r +  |-\frac{P* r^4}{v^2}  |*  \Delta  v]

substituting values

            \Delta  \eta  = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}}  | *  1750   +    | \frac{4 *  4 *10^{5} * (0.002)^3}{0.5*10^{-9}}  |*  0.0002 +  |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2}  |*  0 ]

  \Delta  \eta  = \frac{\pi}{8} [56  +  5120 ]

   \Delta  \eta  = 647 \pi

    \Delta  \eta  = 2032.9

4 0
3 years ago
A chimpanzee sitting against his favorite tree gets up and walks 70.9 m due east and 31.9 m due south to reach a termite mound,
DIA [1.3K]
A right triangle is formed by the 70.9 m walked east and 31.9 m walked south.
The legs of this right triangle are 70.9 and 31.9.
The shortest distance between two points is a straaight line. Therefore the hypotenuse of this triangle is going to be that shortest distance.
We can use the Pythagorean Theorem to find the hypotenuse of this triangle.
a^2+b^2=c^2\\70.9^2+31.9^2=c^2\\5026.81+1017.61=c^2\\6044.42=c^2\\c=\sqrt{6044.42}

c=\sqrt{6044.42}\approx\boxed{77.7458681}\ (decimal\ form)\\\\c=\sqrt{6044.42}=\sqrt{\frac{604442}{100}}=\boxed{\frac{\sqrt{604442}}{10}}\ (exact\ form)

As for the second part of the question, we want to find the angle formed by the hypotenuse and the 31.9 walked east.
We could use any of the three trigonometric ratios here since we know all 3 sides.
sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent
I am going to use tangent, because then I won't have to deal with the hypotenuse and so the answer will be more accurate.

If you haven't already drawn yourself a diagram, now is a good time to.
The side opposite our angle is the 31.9, and the adjacent is 70.9.
Therefore, \tan(m\angle)=\frac{31.9}{70.9}.

We can use inverse trig ratios here to find the measure of our angle.
\tan^{-1}(\tan(m\angle))=\tan^{-1}(\frac{31.9}{70.9})\\\\m\angle=\tan^{-1}(\frac{31.9}{70.9})\approx\boxed{24.2243851\°\ or\ 0.422795279\ rad}
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3 years ago
The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quan
Tpy6a [65]

Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

E_3 =- \frac{13.6 }{3^2}

E_3 =- \frac{13.6 }{9} = -1.511 eV

Energy of the orbit where n = 1 is given as follows

E_1 =- \frac{13.6 }{1^2}

E_1 =- \frac{13.6 }{1} = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

Wavelength in nm = 1244 / energy in eV

= 1244 / 12.089

= 102.90 nm

This represents radiation in ultra-violet region .

8 0
3 years ago
You ride your bike north for 100 m at a constant speed of 5 m/Your acceleration is…
inessss [21]

Answer:

our acceleration is 0 because there is a constant speed..

a = final speed - initial speed / time

final speed = initial speed ( since speed is constant)

therefore acceleration is 0

5 0
2 years ago
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Anika [276]

Answer:

B.

I think.

Explanation:

Mars doesn't have that much of an atmosphere!

Have a great day!

7 0
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