What material reduces thermal energy

How fast must a bug swim to keep up with the waves it produces? How fast must it move to produce a bow wave?

Elden [556K]

Answer:

A bug must swim as fast as the wave speed to keep up with the waves it produces. Moreso, a boat must be moving faster than the waves it creates to produce a bow wave.

6 0

3 years ago

A seagull flying horizontally at 8.00m/s carries a clam with a mass of 300g in its beak. Calculate the total mechanical energy o

Stells [14]

Answer:

9.6J+88.2J=97.8J

Explanation:

Here the velocity of the seagull is given,mass is given and its height.

We have to find its mechanical energy my friend.

Mechanical energy=kinetic energy + potential energy.

First we will find kinetic energy.

For calculating kinetic energy we need mass and velocity,which are given here.

So, Ek=

$1 \div 2mv {?}^{2}$

So by substituting the values we get 9.6J.

Now we find the potential energy which is mgh.

By substituting the values we get 88.2J.

Then we add both of those and get 97.8J

I hope this satisfies you and make sure you contact me if it doesn't

7 0

3 years ago

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$