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Serhud [2]
2 years ago
14

Which statement describes one way in which global winds affect weather and climate? A. Polar easterlies move warm air to the mid

latitudes B. Global winds cause deep ocean currents to form C. In the northern hemisphere, the winds spiral to the left D. The jet stream carries warm air to the north
Physics
2 answers:
Genrish500 [490]2 years ago
4 0

The answer your looking for is "D".

nydimaria [60]2 years ago
4 0

Polar easterlies move cool air to the mid latitudes

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A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of
olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through 90^{o} then position of two mirrors will be changed. The resultant path difference will be as follows.

      \frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}

Formula for change in fringe shift is as follows.

          n = \frac{2lv^{2}}{\lambda c^{2}}

       v^{2} = \frac{n \lambda c^{2}}{2l}

             v = \sqrt{\frac{n \lambda c^{2}}{2l}}

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

             l = 11 m

    \lambda = 5.9 \times 10^{-7} m

           c = 3.0 \times 10^{8} m/s

Hence, putting the given values into the above formula as follows.

            v = \sqrt{\frac{n \lambda c^{2}}{2l}}

               = \sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}

               = 2.41363 \times 10^{9} m/s

Thus, we can conclude that velocity deduced is 2.41363 \times 10^{9} m/s.

3 0
2 years ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0
2 years ago
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
2 years ago
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
Suppose you look out the window of a skyscraper and see someone throw a tomato downward from above your window. your window is a
photoshop1234 [79]
Formula for height 
<span> r(t) = a/2 t² + v₀ t + r₀
</span><span> where 
</span><span> a = acceleration = -32 ft/sec² (gravity) 
</span><span> v₀ = initial velocity 
</span><span> r₀ = initial height 
</span><span> r(t) = -16t² + v₀ t + r₀
</span> <span>Tomato passes window (height = 450 ft) after 2 seconds: 
</span><span> r(2) = 450
</span><span> -16(4) + v₀ (2) + r₀ = 450 
</span><span> r₀ = 450 + 64 - 2v₀ 
</span><span> r₀ = 514 - 2v₀ 
</span><span> Tomato hits the ground (height = 0 ft) after 5 seconds: 
</span><span> r(5) = 0 
</span><span> -16(25) + v₀ (5) + r₀ = 0
</span> r<span>₀ = 16(25) - 5v₀ 
</span><span> r₀ = 400 - 5v₀ 
</span><span> 
 r₀ = 514 - 2v₀ and r₀ = 400 - 5v₀
</span> <span>514 - 2v₀ = 400 - 5v₀
</span><span> 5v₀ - 2v₀ = 400 - 514
</span> <span>3v₀ = −114 
</span><span> v₀ = −38 
</span><span> Initial velocity = −38 ft/sec (so tomato was thrown down) 
</span><span> (initial height = 590 ft) </span>
4 0
2 years ago
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