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Nikitich [7]
3 years ago
13

How are magazine advertisement creating a wrong imperession of young women by doing what?

Physics
1 answer:
Nadya [2.5K]3 years ago
5 0
 These type of commercials lead to little girls, teenagers, and women in general to feel bad about what they look like themselves and have a negative self image. extra anxiety to fit the role of the ultra thin, perfect woman<span> <span>Can lead to individuals engaging in anorexia, bulimia and over-exertion at the gym in order to attain the “perfectly-sculpted” body:
</span></span>
 http://objectificationofads.weebly.com/effects-on-women.html
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State the principle of floatation​
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Explanation:

<em><u>Principle of Floatation</u></em>

Principle of Floatation states that weight of floating body is equal to weight of water displaced by it

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3 years ago
Mangrove forests play an important role in _______.
AysviL [449]
D. All of the above

At high tide fish will feed among the mangrove roots - rich fishing ground

The trees trap sediment and soil in the river that would flow out to sea which also helps stop erosion

Wildlife utilise almost every part of the tree, with insects and birds, monkeys and lizards in the branches, shrimps and fish in the roots, and snails and clams in the soil
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2 years ago
An airplane cabin is pressurized to 570 mmhg. what is the pressure inside the cabin in atmospheres?
abruzzese [7]
1 atm corresponds to 760 mmHg, so we can set up a simple proportion to find how many atmospheres correspond to 570 mmHg:
1 atm: 760 mmHg = x: 570 mmHg
and from this, we find
x= \frac{1 atm \cdot 570 mmHg}{760 mmHg} =0.75 atm
8 0
3 years ago
Read 2 more answers
Which statement best describes how the temperature of the oceans surface water varies
Setler79 [48]
<span>heat capacity→ water has a high heat capacity, and salt water has an even higher one, so the temperatures of the oceans remain within a small range because As the heat rises you get more evaporation which actually cools the ocean down (Specific Heat of Water; Heat needed to break down hydrogen bonds)</span>
8 0
3 years ago
A skier traveling 11.0 m/s reaches the foot of a steady upward 17 â incline and glides 15 m up along this slope before coming to
Darina [25.2K]

The attached free-body diagram shows the forces that are responsible for the skier coming to rest eventually on the incline.

We see that the component of his Weight along the incline mgSin \alpha and the Friction both act in tandem to stop him.

In order to calculate the Friction, we can make use of Newton's 2nd law, which states that F_{net}  = ma

The F_{net} here is given by mgSin \alpha + F_{k}, where F_{k} is the Kinetic Friction.

We also know that the magnitude of Friction Force can be calculated using the equation F_{k} = μ.F_{N}, where F_{N} is the Normal Force acting perpendicular to the incline as shown in the figure.

We see that F_{N}  = mgCos \alpha since they both balance each other out.

Hence, putting all these together, we have mgSin \alpha +μ.mgCos \alpha = ma

Simplifying this, we get gSin \alpha + μgCos \alpha = a

We clearly see that we need to calculate the acceleration before we can obtain the value of the Coefficient of Friction μ

And for that, we make use of the following data obtained from the question:

Initial Velocity V_{i}  = 11.0 m/s

Final Velocity V_{f} = 0

Displacement along the incline D = 15m

Acceleration a = ?

Using the equation V_{f} ^{2}  = V_{i} ^{2}  + 2aD, and

Plugging in known numerical values, we get 0 = (11)^{2} + 2a(15)

Solving for a gives us, a = -4.03 m/s^{2}

Since the negative sign indicates that this is deceleration, we can ignore the sign and consider the magnitude alone.

Thus, plugging in a = 4.03 m/s^{2} in the force equation we wrote above, we have

(9.8)Sin (17) + μ.(9.8)Cos (17) = 4.03

Solving this for μ, we get its value as μ = 0.124

Thus, the average coefficient of friction on the incline is 0.12

6 0
3 years ago
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