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Xelga [282]
3 years ago
10

Can sometimes solve problem 16

Physics
1 answer:
NISA [10]3 years ago
7 0
It’s called the the magnetic field
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Carol can be defined as a social drinker because she can easily limit drinking and she chiefly drinks alcohol when she is out wi
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True. These are the characteristics of a social drinker.
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According to the graph, how many atoms would remain after two half-lives?
Fudgin [204]

Answer:

Let No be initial no of atoms

N = N0 / 2      after 1 half-life

N = N0 / 4     after 2 half-lives

So after 2 half-lives 20 of the 80 atoms remain

4 0
3 years ago
The force of attraction between 2 objects due to their masses is called what?
exis [7]

Answer:

Gravitational force -an attractive force that exists between all objects with mass; an object with mass attracts another object with mass; the magnitude of the force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between the two objects.

pls give brainliest for the answer

4 0
2 years ago
A girl runs 40m due south in 40 seconds an d then 20m due north in 10 secondsCalculate.
m_a_m_a [10]

Answer:

Average speed = 1.2 m/s

Average velocity = 0.4 m/s

Explanation:

Average speed = total distance/total time

Average speed = (40 + 20)/(40 + 10)

Average speed = 60/50

Average speed = 1.2 m/s

Average velocity = displacement/time

Now, she ran 40 m south and ran 20 m back north which is in the direction of where she began the journey.

Thus;

Displacement = 40 - 20 = 20 m

Average velocity = 20/50 = 0.4 m/s

3 0
3 years ago
A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is
hram777 [196]

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0
3 years ago
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