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NARA [144]
3 years ago
14

After the full moon, the moon begins its shrinking or_phases. During these phases, the light side of the moon is on the_

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
7 0
I believe the answer is b
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Dolphins communicate using compression waves (longitudinal waves). Some of the sounds dolphins make are outside the range of hum
lana66690 [7]

Answer: send the message underwater because a more dense medium would make the sound travel faster.

Explanation:

Dolphins communicate using compression waves - longitudinal waves. Longitudinal waves requires a medium to travel. A longitudinal wave transfers energy by the vibration of medium particles in the direction of the wave motion. Compression are the regions where density of the medium is higher and rarefaction is a low density region.

A longitudinal wave travels faster in a denser medium. It has maximum speed in solid and minimum in gas. Thus, to transfer message quickly to dolphin B., dolphin A should send the message underwater and not in air. This is because water has higher density than air. Molecules collide more quickly in water than in air and it takes less time for signal to travel.

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PLS HELP ASAP WILL MARK BRAIN and pls explain your answer....
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What is the definition of differentiation geology?
UkoKoshka [18]
Any process in which a mixture of materials separates out partially
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4 years ago
Identify some of the effects of urbanization
dangina [55]

Answer: Heyyyyyyyyyyyyyy

here you goo

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4 0
4 years ago
A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and
il63 [147K]

Answer:

He becomes a croco-dile food

Explanation:

From the question we are told that

    The height is h = 2.0 m

      The angle is  \theta = 20^o

     The distance is  w = 10m

       The speed is  u = 11 m/s

       The coefficient of static friction is  \mu = 0.02

At equilibrium the forces acting on the motorcycle are mathematically represented as

        ma = mgsin \theta  + F_f

where  F_f is the frictional force mathematically represented as

            F_f =\mu F_x =\mu mgcos \theta

where F_x is the horizontal component of the force

substituting into the equation

            ma = mgsin \theta  + \mu mg cos \theta

            ma  =mg (sin \theta  + \mu cos \theta )

               making  a the subject of the formula

      a = g(sin \theta = \mu cos \theta )

          substituting values

      a = 9.8 (sin(20) + (0.02 ) cos (20 ))

        = 3.54 m/s^2

Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as  

             sin \theta = \frac{h}{l}

making l the subject

          l = \frac{h}{sin \theta }

substituting values

        l = \frac{2}{sin (20)}

           l = 5.85m

Apply Newton equation of motion we can mathematically evaluate the  final velocity at the end of the ramp  as

      v^2 =u^2 + 2 (-a)l

  The negative a means it is moving against gravity

      substituting values

      v^2 = (11)^2 - 2(3.54) (5.85)

           v= \sqrt{79.582}

              = 8.92m/s

The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is

    Initial velocity along the x-axis which is mathematically evaluated as

          v_x = vcos 20^o

       substituting values

         v_x = 8.92 * cos (20)

              = 8.38 m/s

Initial velocity along the y-axis which is mathematically evaluated as

             v_y = vsin\theta

      substituting values

             v_y = 8.90 sin (20)

                  = 3.05 m/s

Now the motion through the pool in the vertical direction can mathematically modeled as

        y = y_o +  u_yt + \frac{1}{2} a_y t^2

where y_o is the initial height,

         u_y is the initial velocity in the y-axis

    a_y  is the  initial  acceleration in the y axis  with a constant value of (g = 9.8 m/s^2)

at the y= 0 which is when the height above ground is zero

      Substituting values

              0 = 2 + (3.05)t - 0.5 (9.8)t^2

The negative sign is because the acceleration is moving against the motion

                 -(4.9)t^2 + (2.79)t + 2m = 0

   Solving using quadratic formula

              \frac{-b \pm \sqrt{b^2 -4ac} }{2a}

substituting values

             \frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}

                t = \frac{-3.05 + 6.9}{-9.8}   \ or t = \frac{-3.05 - 6.9}{-9.8}

                t = -0.39s  \ or  \ t = 1.02s

since in this case time cannot be negative

             t = 1.02s

At this time the position the  motorcycle along the x-axis is mathematically evaluated as

              x = u_x t

               x  =8.38 *1.02

                   x =8.54m

So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of  the pool

7 0
4 years ago
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