Which vector below goes from (0, 0) to (-2, 3)?

How does the Coriolis effect influence the direction of the Trade Winds in the Northern Hemisphere? Does it have the same effect

scoray [572]

Answer:

Part A

Coriolis effect is used to describe how objects which are not fixed to the ground are deflected as they travel over long distances due to the rotation of the Earth relative to the 'linear' motion of the objects

Due to the Coriolis effect the wind flowing towards the Equator from high pressure belts in the subtropical regions in both the Northern and Southern Hemispheres are deflected towards the western direction because the Earth rotates on its axis towards the east

Part B

In the Northern Hemispheres, the winds are known as northeasterly trade winds and in the Southern Hemisphere, they are known as the southeasterly trade wind. Therefore, Coriolis effect has the same effect on the direction of the Trade Winds in the Southern Hemisphere as it does in the Northern Hemisphere

Explanation:

7 0

3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

4 years ago

Which of the following substance is not formed by chemical bonds? A.) Mixture

Rudiy27

The answer to your question would be C

8 0

3 years ago

A key falls from a bridge that is 44 m above the water. It falls directly into a model boat, moving with constant velocity, that

Lady_Fox [76]

The time $t$ it takes for the key to fall 44 m is

$44\,\mathrm m=\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=3.0\,\mathrm s$

(notice I'm taking the downward direction to be positive)

The boat, moving at a presumably constant speed, then has 3.0 s to travel 19 m to the point of impact, which means its speed must be

$v=\dfrac{19\,\mathrm m}{3.0\,\mathrm s}=6.3\,\dfrac{\mathrm m}{\mathrm s}$

6 0

4 years ago

Spaceship 1 and spaceship 2 have equal masses of 300 kg. spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 6 m/s.

Kipish [7]

The final speed is 3 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in absence of external forces, the total momentum of the two spaceships must be conserved. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$