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dimulka [17.4K]

2 years ago

7

An escaped convict runs 1.70 km due East of the prison. He then runs due North to a friend's house. If the magnitude of the conv

ict's total displacement vector is 2.50 km, what is the direction of his total displacement vector with respect to due East?

1 answer:

olga2289 [7]2 years ago

6 0

Answer:

The right approach will be "47° north of east".

Explanation:

The given values are:

East of prison

= 1.70 km

Displacement vector

= 2.50 km

Now,

The direction will be:

⇒ $Cos \ \theta =\frac{1.7}{2.5}$

⇒ $=0.68$

⇒ $\theta=47.16^{\circ}$

i,e., $\theta = 47^{\circ}$ (north of east)

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geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

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2 years ago

Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the block a(t) as a function of time. Express

Inga [223]

Answer:

$a=-\dfrac{k}{m}x(t)$

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From Hooke's law,

$F = -kx(t)$

where $k$ is the spring constant and $x(t)$ is the displacement function measured from the origin. The negative sign indicates the force acts in opposite direction to the displacement. In fact, it is a restoring force; it acts to return the spring to its original undisturbed position.

Since both forces are the same,

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$a=-\dfrac{k}{m}x(t)$

The implication of this is that the acceleration is proportional to the displacement but opposite to it. That last statement is the definition of a simple harmonic motion which this is.

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An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

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- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

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This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

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And re-arranging for F,

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$N sin \theta = m \frac{v^2}{R}$

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galina1969 [7]

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2 years ago