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inna [77]
3 years ago
7

An OSU linebacker of mass 110.0 kg sacks a UM quarterback of mass 85.0 kg. Just after they collide, they are momentarily stuck t

ogether, and both are moving at a speed of 2.60 m/s. If the quarterback was at rest just before he was sacked, how fast was the linebacker moving just before the collision
Physics
1 answer:
expeople1 [14]3 years ago
3 0

Answer:

The initial speed of the linebacker just before the collision was 4.6 m/s.

Explanation:

Mass of OSU linebacker, m = 110 kg

Mass of UM quarterback, m' = 85 kg

Just after they collide, they are momentarily stuck together, the common speed is, V = 2.6 m/s

Initial speed of quarterback is 0 as it was at rest initially, u' = 0

Let u was the initial speed of the linebacker just before the collision. As they struck together, the momentum remains conserved. So,

mu+m'u'=(m+m')V\\\\mu=(m+m')V\\\\u=\dfrac{(m+m')V}{m}\\\\u=\dfrac{(110+85)\times 2.6}{110}\\\\u=4.6\ m/s

So, the initial speed of the linebacker just before the collision was 4.6 m/s.

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STatiana [176]

Answer:

a = 0.55 m / s²

Explanation:

The centripetal acceleration is given by the relation

         a = v² / r

angular and linear velocities are related

         v = w r

we substitute

          a = w² r

In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system

          w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s

let's calculate

          a = 0.105² 50.0

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3 years ago
7. MANSON'S FAVORITE SINGING GROUP WAS...
lukranit [14]
The beach boys i think.. if not im sorry
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3 years ago
How many joules of work are done on an object when a force of 10 N pushes it 5 m?
zhenek [66]

Answer:

option C

Explanation:

given,                            

Force on the object = 10 N

distance of push = 5 m

Work done = ?              

we know,              

work done is equal to Force into displacement.

W = F . s            

W = 10 x 5              

W = 50 J                

Work done by the object when 10 N force is applied is equal to 50 J

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5 0
3 years ago
An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a
prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

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Converting to SI units

10\ ft/s=10\times \dfrac{1}{3.281}

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The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

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3 years ago
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The hiker followed the north trail a distance of two kilometers in thirty minutes is an example that provides a complete scientific description of an object in motion.

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2 years ago
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