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Arte-miy333 [17]
3 years ago
10

At what water depth will a deep-water wave with a wavelength of 40 meters begin to "feel bottom"? 5 meters 15 meters 20 meters 1

0 meters
Physics
1 answer:
ella [17]3 years ago
8 0
The answer is 20 meters would be the water depth started to feel bottom. This is half of the wavelength. The reason behind this is the wave travels half or the partial of the wavelength underneath the surface. For that reason that is where the wave would be starting to feel the ground.
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g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7
bogdanovich [222]

Answer:

When they are approaching each other

    f_a = 2228.7 \  Hz

When they are passing  each other

    f_a = 2100Hz

 When they are retreating  from each other

     f_a =  1980.7 Hz

Explanation:

From the question we are told that

     The velocity of car one is  v_1 = 13.0 m/s

      The velocity of car two is  v_2 = 7.22 m/s

     The frequency of sound from car one is  f_e = 2.10 kHz

Generally the speed of sound at normal temperature is  v = 343 m/s

  Now as the cars move relative to each other doppler effect is created and this  can be represented  mathematically  as

              f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]

Where v_s is the velocity of the source of sound

            v_o is the velocity of the observer of the sound

            f_o is the actual frequence

             f_a  is the apparent frequency

Considering the case when they are approaching each other

        f_a = f_o [\frac{v +  v_o}{v -  v_s} ]

          v_o = v_2  

         v_s = v_1

         f_o = f_e

Substituting value

            f_a = 2100  [\frac{343 +  7.22}{ 343  -  13} ]

              f_a = 2228.7 \  Hz

Considering the case when they are passing  each other    

At that instant

                  v_o = v_s = 0m/s

                   f_o = f_e

               f_a = f_o [\frac{v }{v } ]

              f_a = f_o

Substituting value

             f_a = 2100Hz

Considering the case when they are retreating  from each other    

                f_a = f_o [\frac{v -  v_o}{v +   v_s} ]

          v_o = v_2  

         v_s = v_1

         f_o = f_e      

Substituting value

         f_a = 2100  [\frac{343 -  7.22}{343 +   13} ]    

          f_a =  1980.7 Hz    

7 0
4 years ago
A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra
seraphim [82]

1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

1)

The motion of the bullet is a projectile motion, which consists of two separate motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial speed of the projectile

\theta is the angle of projection

g=9.8 m/s^2 is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

\theta=60^{\circ} (angle)

Solving the equation, we find the horizontal range:

d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m

2)

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

v_y^2 - u_y^2 = 2as

where

v_y is the vertical velocity of the bullet after having covered a vertical displacement of s

u_y is the initial vertical velocity

a =-g= is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

u_y = u sin\theta

Also, the maximum height s is reached when the vertical velocity becomes zero,

v_y =0

Substituting into the equation and re-arranging for s, we find the maximum height:

s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0
4 years ago
A person approaches a flat mirror at a speed of 0.25 m / s. How fast is he approaching his image?
Bezzdna [24]
  • Flat mirror is given here
  • We know that in a flat mirror the distance of object from the mirror is equal to distance of image from the mirror i.e v=u
  • Only the side of image changes ime left is seemed right.

So the speed remains same 0.25m/s.

3 0
3 years ago
A certain mountain road is inclined 3.1° with respect to the horizon. what is the change in altitude of the car as a result of i
BigorU [14]
Given:\\ \alpha =3.1^\circ\\l=2.90km\\\\Find:\\h=?\\\\Solution:\\\\h=l\sin\alpha \\\\h=2.90km\cdot \sin 3.1^\circ=2.90km\cdot 0.054=0,1566km=156.6m
7 0
3 years ago
Sort the processes based on the type of energy transfer they involve.??
klemol [59]
The 1st one goes two added sodoes the second one then the third goes to removed then the fourth goes to added and the rest go to removed
3 0
4 years ago
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