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aleksklad [387]

3 years ago

7

A security guard walks at a steady pace, traveling 190 m in one trip around the perimeter of a building. It takes him 260 s to m

ake this trip.what is his speed. express answer in 2 significant figures

1 answer:

Ber [7]3 years ago

3 0

Answer:

0.7 m/s

Explanation:

The following data were obtained from the question:

Distance travalled (d) = 190 m

Time (t) = 260 secs

Speed (S) =..?

Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:

Speed = Distance (d) /time(t)

S = d/t

With the above formula, we can calculate the speed of the security guard as follow:

Distance travalled (d) = 190 m

Time (t) = 260 secs

Speed (S) =..?

S = d/t

S = 190/260

S = 0.7 m/s

Therefore, the speed of the security guard is 0.7 m/s

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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

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Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

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Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

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$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

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$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

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$\vec F = -5.74 \hat i - 5.22 \hat j$

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Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

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now at t = 1.1 s

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now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

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