Ask question

Ask question

All categories

1 year ago

15

Most of the currents identified have a circular shape. This is because as the air and water move between the equator to the pole

s, they are also affected by the Coriolis effect, which makes them follow a more circular path. Lesson 2.11 Question 10 options: True False

1 answer:

Morgarella [4.7K]1 year ago

5 0

I believe it is true . Somebody correct me if I’m wrong!

You might be interested in

9. All of the following are adaptations of herbivores EXCEPT:

postnew [5]

Answer:

B.KEEN EYESIGHT thats for predators !

Explanation:

Herbivores have multiple stomachs to process tough foods, coats to blend in so they don't get eaten, and tough teeth for chewing on the plants. THEY DO NOT HAVE PREY, THEY ARE THE PREY.

5 0

2 years ago

It requires 3480 J to move a 675 N object how far?

goblinko [34]

5.16 meters!!!!!!!!!!!!!!!!!!!!

4 0

2 years ago

Separating the electron from the proton in a hydrogen atom takes 2.18 ✕ 10−18 j of work. through what electric potential differe

My name is Ann [436]

Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V

6 0

2 years ago

A person throws a ball from height of 6 feet with an initial vertical velocity of 48 feet per second. Use the vertical motion mo

True [87]

Explanation:

Recall the equation for time is distance divided by speed. Here you can use that to solve for "t".

7 0

2 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

(c) $x_f=2.6m$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

(c) Using the kinematics equation:

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

2 years ago