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WARRIOR [948]
3 years ago
12

A drag racer starts her car from rest and accelerates at 11.9 m/s2 for the entire distance of 400 m. What is the speed of the ra

ce car at the end of the run?
Physics
1 answer:
miv72 [106K]3 years ago
6 0
<h2>Speed of the race car at the end of the run is 97.57 m/s</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s  

Acceleration, a = 11.9 m/s²  

Final velocity, v = ?

Displacement,s = 400 m

Substituting  

v² = u² + 2as

v² = 0² + 2 x 11.9 x 400

v = 97.57 m /s

Speed of the race car at the end of the run is 97.57 m/s

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stellarik [79]
<h3>\huge\underline\bold\blue{ƛƝƧƜЄƦ}</h3><h3>Given</h3>

\blue\star v = 20m\s

\blue\star a = 3m\s^2

\blue\star t = 4sec

Firstly we have to find u

\star a = \dfrac{v - u}{t}

\star 3m\s =\dfrac{20 - u}{4}

\star12m\s = 20 - u

\star20 - u = 12m\s

\star- u = -8

\star u = 8

Now we can easily find distance by using second equation of motion

\red\stars = ut + 1\2 at^2

\red\stars = 8(4) + 1\2(3)(16)

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5 0
3 years ago
Consider a plywood square mounted on an axis that is perpendicular to the plane of the square and passes through the center of t
belka [17]

Answer:

T= 8.061N*m

Explanation:

The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:

T = Fr

where F is the force, r the radius.

if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.

Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:

r = \sqrt{(0.38m)^2+(0.38m)^2}

r = 0.5374m

Finally, replacing the value of r and F, we get that the maximun torque is:

T = 15N(0.5374m)

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3 years ago
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0.2289

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Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then

P= 700*9.81*2.5=17167.5 W= 17.1675 kW

To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%

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