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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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Phương trình chuyển động thẳng đều của một chất điểm có dạng: x = 2t – 10. (x: km, t: h). Quãng đường đi được của chất điểm sau
siniylev [52]

Answer:

Distance cover in 2  hour = 6 kilometer

Explanation:

Given equation:

x = 2t - 10

where

x = kilometer

t = hour

Find:

Distance cover in 2  hour

Computation:

T = 2

So,

x = 2t - 10

x = 2(2) - 10

x = 4 - 10

x = -6

Distance cover in 2  hour = 6 kilometer

8 0
3 years ago
Read 2 more answers
If someone were monitoring your vital signs (like your heart rate, oxygen content in your blood, blood pressure, etc.) every hou
Lilit [14]

Answer:

Accuracy

Explanation:

I think accuracy is more important. When it comes to vital organs in the body, the exactness of getting the measurement is paramount. Accuracy deals with getting very close, almost exact you may say, to a known standard. Precision on the other hand, deals with how easy a measurement can be retaken, reproduced or remade, irrespective of how far or close they are from the accepted norm.

From this, we can agree that precision neglects the most important factor, closeness or say, exactness. Precision isn't bothered by it. And while that can be excused in a few instances, it certainly can not be permitted when it comes to life, or organs of the body

7 0
3 years ago
7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del
GrogVix [38]

Answer:

 x =  0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

      sin 60 = T_{y} / T

      T_{y} = T sin 60

       cos 60 = Tₓ / T

      Tₓ = T cos 60

we apply the equation

       ∑ τ = 0

       -W L / 2 - w x + T_{y} L = 0

 

the length of the bar is L = 6m

           -Mg 6/2 - m g x + T sin 60 6 = 0

             x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

             x = (4676 - 5880) / 6860

             x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0
3 years ago
A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of
Paul [167]

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of  pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

<em>h₂ = 348.14 cm</em>

<em>The height of water in the column = 348.14 cm</em>

6 0
3 years ago
Solve for the length of the inclined plane if the angle equals 19.45 degrees.
mel-nik [20]

The length of the inclined plane is approximately 12 ft

The situation forms a right angle triangle.

<h3>Right triangle</h3>

Right triangle have one of its angle as 90 degrees.

Therefore,

The length of the inclined plane is the hypotenuse of the triangle. The length of the inclined plane can be found using trigonometric ratios.

height = 4 ft

angle(∅) = 19.45°

sin 19.45 = 4 / h

h = 4 / 0.33298412235

h = 12.0125847796

h = 12 ft

Therefore, the length of the inclined plane is approximately 12 ft

learn more on inclined plane:brainly.com/question/14163589?referrer=searchResults

5 0
2 years ago
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