Please solve this for 15 points. Please don’t put in a link too!

The California sea lion is capable of making extremely fast, tight turns while swimming underwater. In one study, scientists obs

anygoal [31]

Answer:

Acceleration of Sea Lion is 4.41 g

This is 49% of maximum jet acceleration given as a = 9g

Explanation:

As we know that the radius of the circular loop is given as

R = 0.37 m

The speed of the fish is given as

$v = 4 m/s$

Now the centripetal acceleration of the sea lion is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{4^2}{0.37}$

$a_c = 43.2 m/s^2$

as we know that

$g = 9.8 m/s^2$

so we have

$a = 4.41 g$

Now Percentage of this acceleration wrt maximum jet acceleration is given as

$P = \frac{4.41 g}{9g} \times 100$

$P = 49$ %

6 0

3 years ago

True or false? Magnetic reversals are recorded in the newly formed oceanic crust on BOTH sides of a mid-ocean ridge spreading ce

andrey2020 [161]

answer: false is my answer for thi question

4 0

3 years ago

You hold a metal block of mass 40 kg above your head at a height of 2 m.

Kitty [74]

Answer:

The work done by gravity is 784 J.

Explanation:

Given:

Mass of the block is, $m=40\ kg$

Height to which it is raised is, $h=2\ m$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Now, work done by gravity is equal to the product of force of gravity and the distance moved in the direction of gravity. So,

$\textrm{Work by gravity}=F_g\times h$

Force of gravity is given as the product of mass and acceleration due to gravity.

$\therefore F_g=mg=40\times 9.8=392\ N$ . Now,

$\textrm{Work by gravity}=F_g\times h=392\times 2=784\ J$

Therefore, the work done by gravity is 784 J.

5 0

3 years ago

The amount of energy transported by a wave is most closely related to the wavelength of the wave. True or False?

Artemon [7]

I think its false............ Im pretty sure

3 0

3 years ago

Read 2 more answers

An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of

GaryK [48]

Explanation:

The given data is as follows.

Velocity of bullet, $c_{p}$ = 814.8 m/s

Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

t = $\frac{24.7}{343}$ (velocity in air = 343 m/s)

= 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

$d_{b} = c_{b} \times t$

= $814.8 \times 0.072$

= 58.66

= 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0

3 years ago