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ddd [48]
3 years ago
15

A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later

. How
high is the cliff and how far from the base of the cliff did the diver hit the water?
Help asap
Physics
1 answer:
mart [117]3 years ago
8 0

Explanation:

It is given that,

The horizontal speed of a cliff diver, v_x=3.6\ m/s

It reaches the water below 2.00 s later, t = 2 s

Let d_x is the distance where the diver hit the water. It can be calculated as follows :

d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m

Let d_y is the height of the cliff. It can be calculated using second equation of motion as follows :

d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

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Light from a laser (lambda= 406.192 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen
dsp73

Answer:

The spacing between the slits is    d = 0.00145m                

Explanation:

From the question we are told that

  The wavelength of the light is \lambda = 406.192nm = 406.192*10^{-9} m

   The distance of the slit from the screen is D = 5.937 \ m

    The number of bright fringe is n = 24

     The  length the fringes span is   L = 39.835 mm = \frac{39.835 }{1000} = 0.0398 m

The fringe width (i.e the distance of between two successive bright or dark fringe) is mathematically represented as

             \beta  = \frac{\lambda D}{d}

Where d is  the distance between the  slits

            \beta is the fringe width which can also be evaluated as

                         \beta = \frac{L}{n}

Substituting values

                        \beta = \frac{0.0398}{24}

                          \beta = 1.660 *10^{-3}

Making d the subject of formula in the above equation

                d = \frac{\lambda D}{\beta }

Substituting values

                d = \frac{406.192 *10^{-9} * 5.937 }{1.660 *10^{-3}}

                    d = 0.00145m                

           

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The resistance of a conductor is 10 ohm at 50 degree celsius and 15 ohm at 100 degree celsius. Calculate its resistance at 0 deg
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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

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