A small cart travels 20 meters to the right in 10 seconds calculate the velocity

What is the effect of the author proposing a "thought experiment" at the beginning of the text? FINANCIAL LITERACY

mojhsa [17]

A thought experiment aims to promote speculative thinking, logical reasoning, and paradigm shifts. Thought experiments force us to address problems we find difficult to answer, which forces us out of our comfort zone. They highlight knowledge gaps and aid in recognizing the boundaries of what is knowable.

<h3>What are thought experiments?</h3>

Thought experiments are imaginative methods for examining the nature of things. They are employed for a variety of purposes in a number of disciplines, including physics, economics, history, mathematics, and philosophy.

A thought experiment is a test that we do in our minds. Here, we will imagine a specific scenario, consider some of its repercussions, and come to a broad conclusion.

Therefore, a thought experiment aims to promote speculative thinking, logical reasoning, and paradigm shifts. Thought experiments force us to address problems we find difficult to answer, which forces us out of our comfort zone. They highlight knowledge gaps and aid in recognizing the boundaries of what is knowable.

Learn more about experiments on:

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4 0

1 year ago

Are you an antisocial person. Me honestly am

gregori [183]

Explanation:

By the way.. What's the meaning of ANTISOCIAL??..(ㆁωㆁ)

If you tell me the meaning of it, I'll tell you my answer too(ㆁωㆁ)

7 0

2 years ago

Read 2 more answers

The Moon is responsible for what % of the tides that we experience?

Ivenika [448]

Answer: High tides and low tides are caused by the Moon. The Moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the Moon and the side farthest from the Moon. These bulges of water are high tides.

Explanation:

7 0

3 years ago

Let w(x)=3x-7.If w(x)=14, find x

dlinn [17]

Answer:

7

Explanation:

We are given:

w(x) = 3x - 7

w(x) = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

3x - 7 = 14

3x = 14 + 7

3x = 21

x = 7

So the value of x = 7

6 0

3 years ago

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago