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Taya2010 [7]
3 years ago
15

How does the ocean play an important role in climate and climate change? Keep answer related to the specific heat capacity of wa

ter
Physics
1 answer:
nata0808 [166]3 years ago
3 0
One way that the world's ocean affects weather and climate is by playing an important role in keeping our planet warm. ... The ocean doesn't just store solar radiation; it also helps to distribute heat around the globe. When water molecules are heated, they exchange freely with the air in a process called evaporation.
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The chart shows data for a moving object.
Brut [27]
The object is not accelerating
8 0
3 years ago
Someone please help me answer this !!!
Elan Coil [88]

Answer:

I think its C sorry if it's wrong

7 0
3 years ago
A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th
N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0
3 years ago
A car is traveling with a velocity of 5.5 m/s and has a mass of 1200 kg. What is the kinetic energy?!
vesna_86 [32]

Answer:

<h2>18150 J</h2>

Explanation:

The kinetic energy of the car can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

m is the Mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 1200 \times  {5.5}^{2}  \\  = 600 \times 30.25

We have the final answer as

<h3>18150 J</h3>

Hope this helps you

3 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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