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stellarik [79]
3 years ago
7

Which organisms develop gills from pharyngeal arches and later develop lungs to breathe on land

Physics
2 answers:
mihalych1998 [28]3 years ago
4 0

Answer:

These are the ones you can pick from. I'm on the question and I think it's frog and lizard?

Explanation:

lawyer [7]3 years ago
3 0
Apart from cutaneous respiration<span> present in all </span>species<span>, most lissamphibians are born in an aquatic larval stage with gills. After metamorphosis, they develop lungs to breathe on land. The larvae of urodeles and apods present external, filamentous and highly branched gills which allow them to breathe underwater.
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Which is the equivalent resistance of the circuit<br><br> shown below?
Misha Larkins [42]
1/Rt=1/R1+1/R2+1/R3, 1/Rt=1/3+1/12+1/4=2/3, Rt=equivalent resistance= 1.5 ohms
8 0
3 years ago
If you are going 60 mph what is your speed in m per second
LenKa [72]

Answer:

1/60 mps

Explanation:

We would first have to divide 60 by 60 because there is 60mins per hour to get 1mpm. After that we would have to divide 1 by 60 because there are 60 secs in a min. So our final answer after doing 1/60 would be a fraction.

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2 years ago
Do all objects fall at the same rate?and if they do not why?
Tema [17]
No, I heavier object will fall much faster than something lighter than it. This is because it’s more dense and hard so it can cut through the air particles quicker than a lighter object which takes longer to cut through the air and fall

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Three charges are arranged as shown in the picture above. Find the magnitude and direction of the electrostatic force on the 6 n
bagirrra123 [75]

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6 0
3 years ago
A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

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Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0
3 years ago
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