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3 years ago

Does air pressure increase or decrease with an increase in altitude is the rate of change constant

1 answer:

7 0

The rate of pressure decreases with an increase in altitude is not constant. Rather, pressure decreases rapidly near Earth's surface and more gradually at greater heights.

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Where are you likely to find the oldest oceanic crust?

Stells [14]

Answer:

it would be letter E. near oceanic ridges

Explanation:

new ocean crust is formed at the mid ocean ridges

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3 years ago

In general, metalloids are slightly reactive.

algol [13]

Yes, in general, metalloids are slightly reactive.

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3 years ago

Read 2 more answers

Who showed that our universe is heliocentric—the planets of the solar system revolve around the sun? Johannes Kepler Isaac Newto

denis23 [38]

Answer:

Option (3)

Explanation:

Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.

He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.

This theory describes the present-day motion of the planets around the sun in the solar system. This theory replaced the geocentric theory.

Thus, the correct answer is option (3).

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2 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.