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Gnesinka [82]
3 years ago
12

3 physical adaptations of a frog

Physics
1 answer:
ICE Princess25 [194]3 years ago
6 0
Water, land. breath using skin and lungs
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Throughout the problem, take the speed of sound in air to be 343 .
anyanavicka [17]

Answer:

A pipe has a length of 1.15 m.

a) Determine the frequency of the first harmonic if the pipe is open at each end. The velocity of sound in air is 343 m/s. Answer in units of Hz

b) What is the frequency of the first harmonic if the pipe is closed at one end?

Answer in units of Hz

Explanation:

8 0
3 years ago
When an object is stationary, all of the forces acting on it are balanced
Leviafan [203]
With utmost clarity, that is truest of the truths

5 0
2 years ago
Read 2 more answers
Mike has a mass of 97 kg. He jumps out of a perfectly good airplane that is 2000 m above the ground. After he falls 1000 m, when
Ne4ueva [31]

Answer:

a)  fr = 224.3 N , b)   fr = 224.3 N , c)   v = 198.0  m/s

Explanation:

a) For this exercise let's start by calculating the acceleration in the fall

             v² = v₀² - 2 a (y-y₀)

When it jumps the initial vertical speed is zero

             a = -v² / 2 (y-y₀)

             a = -68 2/2 (1000-2000)

             a = 2,312 m / s²

Let's use the second net law to enter the average friction force

            fr = m a

            fr = 97 2,312

            fr = 224.3 N

b) let's look for acceleration

            v² = v₀² - 2 a y

            a = (v² –v₀²) / 2 (y-y₀)

            a = (4² - 68²) / 2 (0-1000)

            a = 2,304 m / s²

            fr = m a

            fr = 97 2,304

            fr = 223.5 N

c) the speed of the wallet is searched with kinematics

           v² = v₀² - 2 g (y-y₀)

           v = √ (0-2 9.8 (0-2000))

           v = 198.0  m/s

4 0
3 years ago
A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object
Elina [12.6K]

Answer:

2f

Explanation:

The formula for the object - image relationship of thin lens is given as;

1/s + 1/s' = 1/f

Where;

s is object distance from lens

s' is the image distance from the lens

f is the focal length of the lens

Total distance of the object and image from the lens is given as;

d = s + s'

We earlier said that; 1/s + 1/s' = 1/f

Making s' the subject, we have;

s' = sf/(s - f)

Since d = s + s'

Thus;

d = s + (sf/(s - f))

Expanding this, we have;

d = s²/(s - f)

The derivative of this with respect to d gives;

d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²

Equating to zero, we have;

(2s/(s - f)) - s²/(s - f)² = 0

(2s/(s - f)) = s²/(s - f)²

Thus;

2s = s²/(s - f)

s² = 2s(s - f)

s² = 2s² - 2sf

2s² - s² = 2sf

s² = 2sf

s = 2f

8 0
3 years ago
What happens to the appearance of an object as it gets hotter?
lana [24]

Answer:

Explanation:

When an object is heated then it becomes brighter and bluish compared to the initial condition. This happens because when an object is given heat then the electron in the ground state gets excited and reaches some higher state. After reaching a higher state electron make the transition to lower state and simultaneously exhibit the color which is visible with naked eyes.

3 0
2 years ago
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