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2 years ago

7

A model rocket flies horizontally off the edge of a cliff at a velocity of 50.0 m/s. If the canyon below is 100.0 m deep, how fa

r from the edge of the cliff does the model rocket land?

1 answer:

Ugo [173]2 years ago

5 0

First calculate for the time it takes for the model rocket to reach the land. Using the formula:

h = vi t + 0.5 g t^2

where h is height = 100 m, vi is initial vertical velocity = 0 since it simply dropped, t is time = ?, g = 9.8 m/s^2

100 = 0.5 (9.8) t^2

t = 4.52 seconds

So the total horizontal distance taken is:

d = 50 m/s * 4.52 s

<span>d = 225.90 m</span>

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Which of the following are scalar quantities? select all that apply

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Answer:

1, 4, 5, see the explanation below

Explanation:

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2 years ago

A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which

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Answer:

160 Hz , 240 Hz , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

$f_1=\dfrac{320}{4}\ Hz$

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Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

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2 years ago

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you

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PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

$\Delta y = 5 - 1.6 = 3.40 m$

now we will have displacement in horizontal direction

$\Delta x = 14 m$

now we know that

$v_x = vcos54$

$v_y = vsin54$

now we will have

$\Delta x = v_x t$

$14 = (vcos54)t$

also for y direction

$\Delta y = v_y t + \frac{1}{2}at^2$

$3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2$

now from the two equations we will have

$3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2$

$3.40 = 14 tan54 - 4.9 t^2$

$3.40 = 19.3 - 4.9 t^2$

$t = 1.8 s$

now from above equations

$14 = vcos54 (1.8)$

$v = 13.2 m/s$

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

$\Delta y = -1.60 m$

now we will have

$-1.60 = v_y t + \frac{1}{2}at^2$

$-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2$

$4.9 t^2 - 10.7t - 1.60 = 0$

$t = 2.3 s$

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

$t_2 = t - t_1$

$t_2 = 2.3 - 1.8 = 0.5 s$

now the displacement on the other side is given as

$\Delta x = (vcos54) t_2$

$\Delta x = (13.2 cos54)(0.5)$

$\Delta x = 3.88 m$

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2 years ago