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arsen [322]
3 years ago
8

As Merrill watches his finger with both eyes open as he brings his finger closer to his nose, he feels his eye muscles working.

Which depth cue is associated with the feeling he is getting from his eye muscles?
Physics
1 answer:
charle [14.2K]3 years ago
4 0

Answer:

Answered

Explanation:

As Merrill watches his finger with both eyes open as he brings his finger closer to his nose, he feels his eye muscles working. This shows that her eyes  Muscles have both accommodation and convergence.

Accommodation and convergence allows us to view objects both near and at far without double vision.  

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In a step-down transformer, which describes the primary winding compared with the secondary winding? A higher voltage and fewer
kirill115 [55]
B. Higher voltage and more coils of wire :)

7 0
3 years ago
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2. Is there a proportional relationship between voltage and current? If so, write the equation for each run in the form potentia
stepan [7]

Answer:

Yes.

Voltage = resistance×current

Explanation:

There is a proportional relationship between voltage and current. Voltage varies directly as the current provided resistance is constant

Mathematically, voltage = resistance×current where resistance is constant

8 0
3 years ago
A plane is moving due north, directly towards its destination. Its airspeed is 200 mph. A constant breeze is blowing from west t
Vikki [24]

Answer:

200 m/s

Explanation:

Given that,

A plane is moving due north, directly towards its destination. Its airspeed is 200 mph. A constant breeze is blowing from west to east at 60 mph.

We need to find the rate at which the plane is moving towards North. It can be given by :

V_x=v\cos\theta

Where

\theta is the angle with the North

V_x=200\times \cos(\dfrac{60}{200})\\\\=199.99\ m/s\\\\\approx 200\ m/s

Hence, the plane is moving at a rate of 200 m/s.

4 0
3 years ago
What is the name of the perceived change in a sound wave’s frequency due to motion between the observer and the sound source?
Scrat [10]

when observer and source moves relative to each other then the frequency received by the observer is different from the real frequency

This apparent change in frequency due to relative motion is known as Doppler's effect.

Here we know that

f_{app} = f_o\frac{v\pm v_o}{v \pm v_s}

here we know that

f_o = real frequency

v = speed of sound

v_o = speed of observer

v_s = speed of source

so this is known as Doppler's Effect

6 0
4 years ago
Read 2 more answers
A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor
Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

<u>So, the length of the second pipe is L – L’</u>

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm  

Hence, 1.16 cm were cut from the end of the second pipe

6 0
3 years ago
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