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dem82 [27]
3 years ago
9

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

ion of 4.5 m/s squared
Physics
1 answer:
Gekata [30.6K]3 years ago
5 0

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

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A single component that can’t be separated

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4 years ago
A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the
Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$. What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1 $

$x=2 $

and $\rho = 2 \ kg/m^2$ , $P_x=1 \ $ , $P_y=-2 \ $.

So,

$dI = dmr^2$

$dI = \rho \ dA  \ r^2$  ,           $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \  dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

  $= 3050.19 \ kg \ m^2$

So the moment of inertia is  $3050.19 \ kg \ m^2$.

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3 years ago
Lamar writes several equations trying to better understand potential energy. What conclusion is best supported by Lamar’s work?
faust18 [17]
Its B i got it right on the test
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3 years ago
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Harry Potter is chasing his nemesis Draco Malfoy during a quidditch match. Initially, Harry is 35m behind Draco, who has just sp
Sophie [7]

Answer:

the acceleration of harry is equal to 66.126 m/s²

Explanation:

given,

harry is 35 m behind Draco

speed of Draco = 40 m/s

original speed of harry = 50 m/s

acceleration = ?

time taken by the Draco

    t =\dfrac{r}{u} =\dfrac{75}{40}

     t = 1.875 s

distance covered by Harry

  d = 35 + 175 = 210 m

to calculate the acceleration of harry

s = u t+ \dfrac{1}{2}at^2

210 = 50\times 1.875+ \dfrac{1}{2}\times a\times 1.875^2

a × 3.516 × 0.5 = 116.25

a = 66.126 m/s²

hence, the acceleration of harry is equal to 66.126 m/s²

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The flat sheet of paper has more surface area than the crumpled ball
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Read 2 more answers
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