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2 years ago

DOES ANYONE KNOW WAHTS NOISE POLLUSION POEM?

1 answer:

7 0

Noise pollution, also known as environmental noise or sound pollution, is the propagation of noise with ranging impacts on the activity of human or animal life, most of them harmful to a degree.

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Can I also get help on this??

Xelga [282]

Answer:

Explanation:

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2 years ago

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How does state-dependence impact memory recall?

NNADVOKAT [17]

Answer:

The state of the person under the influence of alcohol

The place in which the memory was initially encoded for the retrieval of information.

Retrieval of information is generally better given similar rather than different contextual cues Motivational and emotional factors

These are also likely to affect recall and forgetting.

Explanation:

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2 years ago

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1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown

denis23 [38]

Answer:

a) 3.37 x $10^{3} kg/m^3$

b) 6.42kg/ $m^{3}$

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x $10^{-4}$ $kg/m^{3}$ . So density of metal = mass of metal / volume of metal = 5 / 14 x $10^{-4}$ $kg/m^{3}$ = 3.37 x $10^{3} kg/m^3$

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/ $m^{3}$

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2 years ago

Review. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always st

anyanavicka [17]

To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,

The circumference of the planet is,

$\phi = 25.1m$

Under the mathematical value the radius would be

$\phi = 2\pi r$

$r = \frac{25}{2\pi}$

$r = 3.9788km$

Using second equation of motion

$x = \frac{1}{2} at^2$

Replacing the values given,

$1.4 = \frac{1}{2} a (29.2)^2$

Rearranging and solving for 'a' we have,

$a = 0.003283m/s^2$

Using the value of acceleration due to gravity from Newton's law we have that

$a = \frac{GM}{r^2}$

Here,

r = Radius of the planet

G = Gravitational Universal constant

M = Mass of the Planet

$\frac{(6.67*10^{-11})*M}{(3.9788*10^3)^2} = 0.003283$

$M = 7.79201*10^{14}kg$

Therefore the mass of this planet is $7.79201*10^{14}kg$

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3 years ago

How do we move in space without any friction?

Luba_88 [7]

Answer:

The Physical Behavior of Objects when Gravity is Missing

In order to be able to form a concept of the general physical conditions existing in a weightless state, the following must be noted: the force of the Earth's gravity pulling all masses down to the ground and thus ordering them according to a certain regularity is no longer active.

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3 years ago