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Serhud [2]
2 years ago
14

Which statement describes one way in which global winds affect weather and climate? A. Polar easterlies move warm air to the mid

latitudes B. Global winds cause deep ocean currents to form C. In the northern hemisphere, the winds spiral to the left D. The jet stream carries warm air to the north
Physics
2 answers:
Genrish500 [490]2 years ago
4 0

The answer your looking for is "D".

nydimaria [60]2 years ago
4 0

Polar easterlies move cool air to the mid latitudes

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How are adhesion and cohesion similar? how are they different?
hichkok12 [17]
<span> they're both properties of water. both involve the "sticking" property of H2O .</span><span> cohesion is "sticking" together (water sticking to water) and adhesios is water</span><span>
</span> "sticking" to something else. They are both important to life processes.

 <span>In chemistry, adhesion refers to the tendency of unlike molecules to bond with one another, while cohesion refers to the attractive force between molecules of the same type. Surface tension, which is an essential property of water, illustrates the relationship between adhesion and cohesion.</span>
4 0
2 years ago
Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter
satela [25.4K]

The time-period of a simple pendulum is

<em>Time =  2 π √(length/grav-accel)</em>

After unraveling the question, then completing it, and working out what I <em>believe</em> it's trying to ask, the choice that correctly answers the question that I have invented is <em>choice-E</em> .

7 0
1 year ago
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
1 year ago
Yolanda has decided she is going to be a nurse. What is the next step she should take to make her career goals a reality? a. Cre
andrew11 [14]

The correct answer is option D

When someone decides to do something then he or she must prepare themselves according to the career they choose.

When Yolanda has decided to be a nurse then she must create an individual career plan and start working on it.

A goal oriented person only works in the area of their goal and accomplishes it.


8 0
1 year ago
Read 3 more answers
In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric
Viktor [21]

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

Comet mass M = 1.0*10^{13} kg

Radius r = 1.6km = 1600 m

Rock was dropped from a height 'h' from surface = 1m

The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

g = \frac{GM}{R^2}

Where G means gravitational universal constant and M the mass of the planet

g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}

g = 2.607*10^{-4} m/s^2

Now calculate the value of the time

h = \frac{1}{2} gt^2

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}

t = 87.58s

The time taken for the rock to reach the surface is t = 87.58s

8 0
1 year ago
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