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3 years ago

Which ancient society developed the cuneiform writing system

Physics

2 answers:

Anuta_ua [19.1K]3 years ago

5 0

The correct answer is B

yulyashka [42]3 years ago

5 0

The first ancient society to develop cuneiform writing where the Sumerians who where part of Mesopotamia so the answer to this question would be B hope my answer helped :)

You might be interested in

How does mass and material type effect thermal energy transfer

Vedmedyk [2.9K]

The mass contributes with the time of thermal energy transfer with respect to the material type but most importantly the material type will determine rate at which the material absorbs the transfer of heat or thermal energy by either three types, conduction, convection and radiation.

4 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

Two atoms that are isotopes of one another must have the same number of what?

fredd [130]

Answer:

Explanation:

Two atoms which are isotopes of one another must have a different number of neutrons.

Isotopes are defined as atoms of the same element which have the same numbers of protons i.e. atomic number remains the same, but has different numbers of neutrons. It is observed that they have same chemical properties due to the same electronic configuration but physical properties differs.

3 0

3 years ago

The of the line on a velocity vs. time graph represents acceleration.

Eva8 [605]

Answer:

~The slope of the line on a velocity vs. time graph represents acceleration.

Explanation:

~~Acceleration is equal to the ratio between the change in velocity of an object and the time taken:

a=\frac{\Delta v}{\Delta t}a=

Δt

Δv

On a velocity-time graph, this ratio corresponds to the slope of the line. In fact, \Delta vΔv corresponds to the increment in the y-value (the velocity), while \Delta tΔt corresponds to the increment in the x-value (the time), therefore their ratio corresponds to the definition of slope of the line.

5 0

3 years ago

A puppy is running with a speed of 11.2 m/s towards it owner when it

Levart [38]

Answer:

-5.6m/s²

Explanation:

Given parameters:

Initial speed = 11.2m/s

Time of running = 2s

Unknown:

Acceleration of the puppy = ?

Solution:

Acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{Final speed - initial speed }{time}$

final speed = 0m/s

Acceleration = $\frac{0 - 11.2}{2}$ = -5.6m/s²

8 0

3 years ago