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STALIN [3.7K]
3 years ago
9

Which best describes the definition for the atomic mass of an element

Physics
2 answers:
Verizon [17]3 years ago
7 0

Answer:

Atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes. - Mass number: count of the total number of protons and neutrons in an atom's nucleus.

dezoksy [38]3 years ago
5 0

Answer:

it is a weighted average of the masses of an element’s isotopes

Explanation:

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The position x, in meters, of an object is given by the equation:
LenaWriter [7]

Answer:

The SI units of A, B and C are :

m,\ m/s\ and\ m/s^2                  

Explanation:

The position x, in meters, of an object is given by the equation:

x=A+Bt+Ct^2

Where

t is time in seconds

We know that the unit of x is meters, such that the units of A, Bt and Ct^2 must be meters. So,

  • A=m
  • bt=m

b=\dfrac{m}{s}=m/s

  • Ct^2=m

C=m/s^2

So, the SI units of A, B and C are :

m,\ m/s\ and\ m/s^2

So, the correct option is (B).

3 0
3 years ago
1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic​
Korvikt [17]

51.448 g is the required answer!

8 0
3 years ago
If a magnet is broken into two pieces, what happens to the magnetic poles?
Nostrana [21]

Answer:

The second is correct.

Explanation:

Try with a magnet and experiment. You'll find out

6 0
2 years ago
Read 2 more answers
0. A 3.00-kg block is dropped from rest on a vertical spring whose spring constant is 750 N/m. The block hits the spring, compre
Dmitry_Shevchenko [17]

Answer:

the spring compressed is 0.1878 m

Explanation:

Given data

mass = 3 kg

spring constant k = 750 N/m

vertical distance h = 0.45

to find out

How far is the spring compressed

solution

we will apply here law of mass of conservation

i.e

gravitational potential energy loss = gain of eastic potential energy of spring

so we say m×g×h = 1/2× k × e²

so e² = 2×m×g×h / k

so

we put all value here

e² = 2×m×g×h / k

e² = 2×3×9.81×0.45 / 750

e²  = 0.0353

e = 0.1878 m

so the spring compressed is 0.1878 m

5 0
3 years ago
Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

3 0
3 years ago
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