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KIM [24]
4 years ago
15

PLEASE HELLPP ILL GIVE BRAINLY

Physics
1 answer:
AlexFokin [52]4 years ago
3 0

Answer:

B ( I think so?)

Explanation:

Ocean currents are driven by wind, water density differences, and tides. Oceanic currents describe the movement of water from one location to another.

Coastal currents are intricately tied to winds, waves, and land formations. Winds that blow along the shoreline—longshore winds—affect waves and, therefore, currents.

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Describe three ways that adults continue to guide children's emotions during the school-age
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1. Always be the bigger person
2. Violence is never the answer
3. Don’t fight fire with fire
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3 years ago
What are the different types of precipitation
nikitadnepr [17]

There are many different types of precipitation —rain, snow, hail, and sleet for example—yet they all have a few things in common. They all come from clouds. They are all forms of water that fall from the sky.

8 0
3 years ago
A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm b
Artist 52 [7]

<u>Answer:</u>

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

<u>Explanation:</u>

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

    Distance traveled during time t = 4 cm

    Initial velocity = u m/s

   Acceleration = 0 m/s^2

So 4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}

Now at time 2t,

  S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm

  So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

  Distance traveled during time t = 4 cm

    Initial velocity = 0 m/s

   Acceleration = -g m/s^2

4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}

At time 2t,

     S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm

 So vertical distance traveled in time 2t = 16 cm to the bottom

 

6 0
3 years ago
A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari
ElenaW [278]

Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency  c/λ  where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

7 0
3 years ago
The diagram illustrates the movement of sound waves between an observer and a race car. As the race car drives away from the obs
Andrei [34K]
I believe the answer is d
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3 years ago
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