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2 years ago

Determine the right answer to fill in the blank spaces

Physics

1 answer:

ratelena [41]2 years ago

5 0

Answer:

Explanation:

A pure substance is something that is entirely made up of particles that are identical to each other.

Any substance that is not pure, must be a mixture.

We are surrounded by mixtures. The air is a mixture of gases . The oceans are a mixture of (mainly) water and salt. The solid earth is mostly rock, which is a mixture of different minerals.

Natural resources are substances we need and use, which occur naturally. Some come from living things, (example) cotton other are non-living (example) sand.

The opposite of a natural resource is a made resource.

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When water freezes, its volume increases by 9.05%. What force per unit area is water capable of exerting on a container when it

AnnZ [28]

Answer:

The Pressure is 0.20 MPa.

(b) is correct option.

Explanation:

Given that,

Change in volume = 9.05%

{tex]\dfrac{\Delta V}{V_{0}}=0.0905[/tex]

We know that.

The bulk modulus for water

$B=0.20\times10^{10}\ N/m^2$

We need to calculate the pressure difference

Using formula bulk modulus formula

$B=\Delta P\dfrac{V_{0}}{\Delta V}$

$\Delta P=B\dfrac{\Delta V}{V_{0}}$

$\Delta P=0.2\times10^{10}\times0.0905$

$\Delta P=0.2\times10^{6}\ Pa$

$\Delta P=0.20 MPa$

Hence, The Pressure is 0.20 MPa.

6 0

4 years ago

Photographs of many young stars show long jets of material apparently being ejected from their poles.

Rudiy27

Answer:

A) True

Explanation:

Researchers have detected numerous jets of gas ejected from poles of young stars and planetary nebulae.

By examining images of hydrogen molecules excited at infrared wavelengths, scientists have been able to see through the gas and dust in the Milky Way, in order to observe the most distant targets. These goals are normally hidden from view and many of them have never been seen before.

The entire study area covers approximately 1,450 times the size of the full moon, or the equivalent of an image of 95 gigapixels. The survey reveals jets emanating from proto-stars and planetary nebulas, as well as remnants of supernovae, the illuminated edges of vast clouds of gas and dust, and the warm regions that surround massive stars and their associated groups of smaller stars.

7 0

3 years ago

An alien from the newly discovers planet nine weighs himself on his planet and finds his weight to be 3200 N. When he stoves on

White raven [17]

Answer:

2.45 m/s²

Explanation:

From the question,

On the Earth

W = mg.................. Equation 1

Where W = weight of the alien on the earth, m = mass of the alien on the earth, g = acceleration due to gravity of the earth.

Make m the subject of equation 1

m = W/g................... Equation 2

Given: W = 3200 N

Constant: 9.8 m/s²

Substitute these value into equation 2

m = 3200/9.8

m = 326.5 kg.

Similarly,

On planet 9,

W' = mg'............... Equation 3

Where W' = weight of the alien on planet 9, g' = acceleration due to gravity on planet 9.

make g' the subject of the equation

g' = W'/m............ Equation 4

Given: W' = 800 N

Substitute into equation 4

g' = 800/326.5

g' = 2.45 m/s²

5 0

3 years ago

A 3kg object moving at 4 m/s encounters a 20 N resistive force over a duration of 0.20 seconds. The impulse experienced by this

WITCHER [35]

Answer:

the impulse experienced by this object is 4 Ns

Explanation:

Given;

mass of the object, m = 3 kg

velocity of the object, v = 4 m/s

resistive force, F = 20 N

duration of impact, t = 0.2 s

The impulse experienced by this object is calculated as follows;

J = F x t

J = 20 x 0.2

J = 4 Ns

Therefore, the impulse experienced by this object is 4 Ns

3 0

3 years ago

One particle has a charge of 2.15 x 10^ -9 while another particle has a charge of 3.22 x 10^ -9 If the two particles are separat

marin [14]

Answer:

B. 2.77 x $10^{-4}$ N

Explanation:

The required force can be calculated by:

F = $\frac{Kq_{1}q_{2} }{d^{2} }$

Where F is the force between the particles, K is the coulomb's constant (9 x $10^{9}$ $Nm^{2}/C^{2}$ ), $q_{1}$ is the charge on the first particle, $q_{2}$ is the charge on the second particle and $d^{2}$ is the distance between the particles.

So that:

F = $\frac{9*10^{9}*2.15*10^{-9} *3.22*10^{-9} }{(0.015)^{2} }$

= $\frac{6.2307*10^{-8} }{(2.25*10^{-4} }$

= 2.7692 x $10^{-4}$

The force between the particles is 2.77 x $10^{-4}$ N.

3 0

3 years ago