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3 years ago

You leave your home at 1pm. At 3pm, you are 100 km east of your house. What was your average velocity in km/hr

Physics

1 answer:

Rasek [7]3 years ago

4 0

Answer:

50km/h

Explanation:

Average velocity = change in distance (or distance travelled) divided by/ the change in time (or time taken.)

The change in distance has been given as 100km.

The change in time is 3pm-1pm = 2 hours.

Therefore the average velocity was 100/2 = 50km/h (to the east).

Hope this helped!

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Pani-rosa [81]

<span>Even though the Sun has a greater mass than Earth, the Moon orbits Earth because it's closer to the Earth than to the Sun. Because of this proximity between the Earth and the Moon, the Earth has a stronger gravitational pull than the Sun does. Furthermore, the Earth's mass is 81 times that of the Moon, and so at this proximity, it is more than able to overpower what pull the Sun exerts on the Moon.</span>

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3 years ago

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a student calculates experimentally the value of density of an iron as 4.4 gcm³. if the actual density of an iron is 7.6 gcm³, c

ser-zykov [4K]

Hola!

Percentage Error is a measurement of the discrepancy between an observed and a true, or accepted value.

[ refer the attachment. ]

According to Question,

% error = $\frac{7.4 - 7.6}{7.6}$ × 100

= 2.631 % = 2.7 % (approximately.)

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3 years ago

What is a central idea

Mice21 [21]

Answer:

The definitive and unifying theme or idea of a story or article. It encompasses all the aspects necessary to create a coherent main idea. The central idea is typically expressed as a universal truth or theme that is built and supported by the setting and characters in a story.

Explanation:

This is the correct answer to your question.

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Kyle.

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3 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.