Winds are named for the cardinal direction they blow from. Hence, a wind that "blows towards the east", logically should come from the west and is called a "west wind".

In thise sense, one of the best examples of this type of wind are the Westerlies, which are are prevailing winds that blow from the west at midlatitudes and have the characteristic that are stronger during winter and weaker during summer.

Therefore, the statement is false.

6 0

2 years ago

A 100-N force causes an object to accelerate at 2 m/s/s. What is the mass of the object?

Kitty [74]

Answer:

$50\; \rm kg$ .

Explanation:

By Newton's Second Law, the acceleration $a$ of an object is proportional to the net force $\sum F$ on it. In particular, if the mass of the object is $m$ , then

$\sum F = m \cdot a$ .

Rewrite this equation to obtain:

$\displaystyle m = \frac{\sum F}{a}$ .

In this case, the assumption is that the $100\; \rm N$ force is the only force that is acting on the object. Hence, the net force $\sum F$ on the object would also be

Make sure that all values are in their standard units. Forces should be in Newtons (same as $\rm kg \cdot m \cdot s^{-2}$ , and the acceleration of the object should be in meters-per-second-squared ( $\rm m \cdot s^{-2}$ ). Apply the equation $\displaystyle m = \frac{\sum F}{a}$ to find the mass of the object.

$\displaystyle m = \frac{100\; \rm N}{2\; \rm m \cdot s^{-2}} = 50\; \rm kg$ .

4 0

3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago