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Assoli18 [71]
3 years ago
5

What are tadpoles? How does it look like

Physics
2 answers:
Bingel [31]3 years ago
7 0

Answer:

Tadpoles are baby frogs. The tadpole looks like a frog with a tail and becomes a froglet around 12 weeks after hatching. The tadpole becomes an adult frog around 13 to 16 weeks after hatching.

Explanation:

I also attached a photo for you

Brainliest or rating please and thank you

Download pdf
AnnZ [28]3 years ago
6 0

Answer: a tadpole is the first or second step into the evolution for a frog

Explanation: a frog starts with the egg then it becomes a tadpole for 6 to 12 weeks before it can start going out the water.

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What happens when a person’s immune system is very weak?
Alexeev081 [22]
The correct answer is C) becuse without certain medicine they will dieeeee
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3 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
What do an earthquake wave and a light wave have in common?
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It’s c.) I think so at least
5 0
3 years ago
Read 2 more answers
A window in a skyscraper has a surface area of 3.50 m^2. Wind rushes by the outside of the window at 17.4 m/s, while inside the
Mariana [72]

The difference in the pressure between the inside and outside will be 369.36 N/m²

<h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

The given data in the problem is;

dP is the change in the presure=?

Using Bernoulli's Theorem;

\rm  \rho\frac{V^2_{12}}{2} +P_1= \rho \frac{V^2_{22}}{2} +P_2 \\\\\ P_2-P_1=\rho \frac{v_2^2-v_1^2}{2} \\\\  P_2-P_1= 1.21 \times \frac{17.4^2-0}{2} \\\\  \triangle p=369.36 \ N/m^2

Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²

To learn more about the pressure refer to the link;

brainly.com/question/356585

#SPJ1

3 0
2 years ago
In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys
nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

\Delta U=q+w\\\\U_{final}-U_{initial}=q+w

where,

U_{final} = final internal energy

U_{initial} = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and U_{initial}

initial internal energy = U_{initial} = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

U_{final}-(2000J)=(1000J)+(500J)

(4) By solving the terms, we get

U_{final}-(2000J)=(1000J)+(500J)

U_{final}-(2000J)=1500J

U_{final}=2000J+1500J

U_{final}=3500J

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

5 0
3 years ago
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