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anygoal [31]
3 years ago
15

A stone is thrown horizontally with an initial speed of 9 m/s from the edge of a cliff. A stop watch measures the stone's trajec

tory time from the top of the cliff to the bottom to be 4.7 s. What is the height of the cliff? Round to the nearest tenth of a meter
Physics
1 answer:
Nady [450]3 years ago
6 0

Answer:

the height of the cliff is equal to 108.241 m

Explanation:

given,

initial speed of the stone horizontally  = 9 m/s

vertical speed of the stone = 0 m/s

time taken of the trajectory = 4.7 s

using equation of motion

s = ut + \dfrac{1}{2}at^2

s =0 \times 4.7 + \dfrac{1}{2} \times 9.8 \times 4.7^2

s = 0 + 4.9 \times 4.7^2

s = 4.9 \times 22.09

s = 108.241 m

hence, the height of the cliff is equal to 108.241 m

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Matter can exist in one of three main states: solid, liquid, or gas. Solid matter is composed of tightly packed particles. A solid will retain its shape; the particles are not free to move around. Liquid matter is made of more loosely packed particles. Hopefully this helps:)

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3 years ago
Read 2 more answers
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
Statistically, consumers tell more people about negative experiences in business encounters than about positive experiences. A)
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true hope this helps


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3 years ago
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A rescue plane flying horizontally at 72.6 m / s spots a survivor in the ocean 182 m directly below and releases an emergency ki
Mila [183]

Answer:

547 m

Explanation:

From law of motion

s = ut + ½at²

Where "t" is Time taken to reach Earth

s= distance= 182 m

a= vertical acceleration = 5.82 m / s 2

U= initial velocity in vertical position = 0

182= ½ × 5.82t²

t²=( 2× 182)/ 5.82

= 364/5.82

= 62.54

t= √62.54

t= 7.908s

horizontal distance travelled = speed x time

Horizontal speed= 72.6 m / s

horizontal distance travelled =72.6× 7.908

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Hence, the survivor will it hit the waves at 547 m away

3 0
2 years ago
When you changed from low to high power, how did the change affect the working distance of the lens?
Basile [38]

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about  working distance

brainly.com/question/13551539

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