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Serga [27]
3 years ago
8

An element is ____

Physics
2 answers:
Likurg_2 [28]3 years ago
8 0

An element is C. A substance with only one type of Atom

Arisa [49]3 years ago
5 0

Answer:

C. A substance with only one type of atom

Explanation:

An element is a type of substance which is made up of only one type of an atom. Taking the example the hydrogen is made up of 1 proton and 1 electron. If any one look at the atom of hydrogen then he notice that the most of the atoms of hydrogen do not contain any neutrons but few of them contain 1 or 2 neutrons.

These different type of hydrogen is known as isotopes which contains same no. of protons but different no. of neutrons.

If you change the proton which means you change the element but if you change the neutron no. means you change its isotope.

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VashaNatasha [74]
Two half lives so it is 4000 years old
5 0
3 years ago
Read 2 more answers
The magnetic lines of force always travel from north to south true or false
kumpel [21]

Answer:

True

Explanation:

Magnetic field lines outside of a permanent magnet always run from the north magnetic pole to the south magnetic pole. Therefore, the magnetic field lines of the earth run from the southern geographic hemisphere towards the northern geographic hemisphere.

7 0
1 year ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
What is the distance an object would be from Earth if its parallax were one arcsecond?
ZanzabumX [31]
The correct answer is (a.) a parsec. A parsec is a distance an object would be from Earth if its parallax were one arcsecond. This unit of measurement is usually used in astronomy which makes it easier for astronomers to calculate or measure in space accurately. 
8 0
3 years ago
If a person weighs 818 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that
alexandr402 [8]

Answer:

g'=63.74\ m/s^2

Explanation:

It is given that,

Weight of the person on Earth, W = 818 N

Weight of a person is given by the following formula as :

W=mg

g is the acceleration due to gravity on earth

m=\dfrac{W}{g}

m=\dfrac{818\ N}{9.8\ m/s^2}

m = 83.46 kg

The mass of an object is same everywhere. It does not depend on the location.

Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N

g' is the acceleration due to gravity on that planet. So,

g'=\dfrac{W'}{m}

g'=\dfrac{5320\ N}{83.46\ kg}                

g'=63.74\ m/s^2

So, the acceleration due to gravity on that planet is 63.74\ m/s^2. Hence, this is the required solution.                                                                    

6 0
3 years ago
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