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SOVA2 [1]
2 years ago
6

6. A wave has a frequency of 600 Hz and is traveling at 300 m/s. What is its wavelength?

Physics
1 answer:
Luda [366]2 years ago
5 0

Answer:

0.5m

Explanation:

v=f×lamda

v is 300m/s, f is 600Hz, lamda is ?

lamda=v/f

lamda=300/600

lamda =3/6=1/2m

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An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
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Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

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Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

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b) The horizontal component will not change it remains same as 41 m/s

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          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

4 0
3 years ago
A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius
ahrayia [7]

Answer:

The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.

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E = 0.

at R < r < 2R:

The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.

at 2R < r:

The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.

Explanation:

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2 years ago
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