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3 years ago

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An Alaskan rescue plane traveling 45 m/sdrops a package of emergency rations froma height of 145 m to a stranded party of reese

(enr647) – Homework 3, 2d motion 19-20 – dowd – (McdanielMPHY119203)3explorers.The acceleration of gravity is 9.8 m/s2.Where does the package strike the groundrelative to the point directly below where it

1 answer:

Artist 52 [7]3 years ago

4 0

Answer:

package will strike at a distance of 244.7 m

Explanation:

Given data;

Velocity of plane 45 m/s

Acceleration due to gravity 9.8 m/s^2

horizontal component of plane velocity is ux = 45 m/s

vertical component is zero

in the same way

ay = - 9.8 m/s^2

ax = .0 m/s^2

motion in y direction = -145 m

from equation of motion time of flight is

$s = ut + \frac{1}{2} at^2$

$-145 = 0 + \frac{1}{2} (-9.8) t^2$

solving for t we get

t = 5.439 sec

Motion in x direction

$s = ut + \frac{1}{2} at^2$

$= (45 \times 5.43) + \frac{1}{2} (0) 5.43^2$

s = 244.79 m

Therefore package will strike at a distance of 244.7 m

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A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

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3 years ago

Mark and David are loading identical cement blocks onto David’s pickup truck. Mark lifts his block straight up from the ground t

Pepsi [2]

Answer:

b) true. The jobs are equal

Explanation:

The work on a body is the scalar product of the force applied by the distance traveled.

W = F. d

Work is a scalar, the work equation can be developed

W = F d cos θ

Where θ is the angle between force and displacement

Let's apply these conditions to the exercise

a) False, if we see the expression d cosT is the projection of the displacement in the direction of the force, so there may be several displacement, but its projection is always the same

b) true. The jobs are equal dx = d cosθ

c) False, because the force is equal and the projection of displacement is the same

d) False, knowledge of T is not necessary because the projection of displacement is always the same

e) False mass is not in the definition of work

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3 years ago

A ball is thrown up at a speed of 20 m/s.

ioda

Given:

(Initial velocity)u=20 m/s

At the maximum height the final velocity of the ball is 0.

Also since it is a free falling object the acceleration acting on the ball is due to gravity g.

Thus a=- 9.8 m/s^2

Now consider the equation

v^2-u^2= 2as

Where v is the final velocity which is measured in m/s

Where u is the initial velocity which is measured in m/s

a is the acceleration due to gravity measured in m/s^2

s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.

Substituting the given values in the above formula we get

0-(20x20)= 2 x- 9.8 x s

s= 400/19.6= 20.41m

Thus the maximum height attained is 20.41 m by the ball

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3 years ago

A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.

scoundrel [369]

Answer:OB=58.3m

Explanation:

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now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB

= $\sqrt{[37.084]^{2}+[44.984]^{2} }$

= $\sqrt{3398.78}$

OB = $58.3$

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2 years ago