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3 years ago

Mesopotamians also were the first to use ___________.

Physics

2 answers:

rusak2 [61]3 years ago

7 0

I think it may be c i learned about this last year

Bumek [7]3 years ago

5 0

Answer:

Explanation:

The Sumerians were very inventive people. It is believed that they invented the sailboat, the chariot, the wheel, the plow, and metallurgy.

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zloy xaker [14]

Answer: Can't see the photo clearly.

6 0

2 years ago

How many calories are absorbed by a pot of water with a mass of 500g in order to raise the temperature from 20c to 30c?

LenaWriter [7]

Calories would denote the amount of heat.

Givens are:
Mass = 500 g
t0 = 20C
tf = 30C
C = 1 cal/gC

Formula:
Q=MCt

500g (1cal/gC) 10C= 5000 cal

Thank you for your question. Please don't hesitate to ask in Brainly your queries.

8 0

3 years ago

How are Newton’s second and third laws of motion important to your everyday life?

Rasek [7]

Well, first off, Newtons second law of motion deals with the motion of accelerating and decelerating objects.
We already know that from everyday life examples such as simply pushing a car that if 2 people push a car on a flat road it will accelerate faster than if one person was pushing it... Therefore, there is a relationship between the size of the force and the acceleration.
Now onto the third law of motion. First of all, what is the third law of motion? Well, a force is a push or a pull that acts upon an object as a results of its interaction with another object. Forces result from interactions! According to Newtons third law, whenever one object, and another object interact with each other, they exert forces upon each other. "For every action, there is an equal and opposite reaction." The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. So, how is this important to everyday life you may ask?
Well, the action-reaction force pairs are found everywhere in your body.
For example, right now as I am typing, my tendons are exerting forces on bones, and those bones exert reaction forces on the tendons, as muscles contract, pulling my fingers on the keys. I press on those keys, and they press back on my fingers. See? Since i'm pressing on the keys, the press back on me. Its opposite from each other, as stated in the quite above. "For every action, there is an equal and opposite reaction."

3 0

4 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago

A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no

frosja888 [35]

Answer:

Explanation:

Given a circular loop of radius R

r = R

Note: the radius lies in the xy plane

Area is given as

A = πr² = πR²

At t = 0, no magnetic field B=0

The magnetic field is given as a function of time

B = C•exp(t) •i + D•t² •k

Where C and D are constant

We want to find the magnitude of EMF in the circular loop.

EMF is given as

ε = - N•dΦ/dt

Where,

N is number of turn and in this case we will assume N = 1.

Φ is magnetic flux and it is given as

Φ = BA

ε = - N•d(BA)/dt

Where A is a constant, then we have

ε = - N•A•dB/dt

B = C•exp(t) •i + D•t² •k

dB/dt = C•exp(t) •i + 2D•t •k

Then,

ε = - N•A•dB/dt

ε = - 1•πR²•(C•exp(t) •i + 2D•t •k)

ε = -πR²•(C•exp(t) •i + 2D•t •k)

So, let find the magnitude of EMF

Generally finding magnitude of two vectors R = a•i + b•j

Then, |R| = √a² + b²

So, applying this we have,

ε = πR² (√(C²•exp(2t) + 4D²t²))

From the given magnetic field, we are given that,

B = 0 at t = 0

B = C•exp(t) •i + D•t² •k

B = 0 = C•exp(0) •i + D•0² •k

0 = C

Then, C = 0.

So, substituting this into the EMF.

ε = πR² (√(0²•exp(2t) + 4D²t²))

ε = πR² (√4D²t²)

ε = πR² × 2Dt

ε = 2πDR²t

So, the EMF is also a function of time

ε = 2πDR²t

4 0

4 years ago