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1 year ago

If air resistance acts on a falling object will all of its potential energy be converted into kinetic energy?

Physics

1 answer:

Mademuasel [1]1 year ago

6 0

Answer:

No, some energy will be dissipated energy due to work of air resistance.

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In what ways can students use time management? Check all that apply.

Korvikt [17]

Answer: the answer is A, C, E

Explanation:

Time management is apart of your study schedule so you will need to determine when to study, same goes with how to use ur study time, lastly it helps create ur study schedule to help find out what time you are free to study. :)

7 0

2 years ago

Suppose your local apparent solar time is 11 am, and you have a clock that says it is noon in Greenwich. What is your longitude

jeyben [28]

The longitude based on the time difference is 15 degrees.

<h3>Longitude of complete rotation of the Earth</h3>

The longitude of a complete rotation of the earth in a 24 hours is calculated as follows;

$= \frac{360^0}{24} = 15^0 \ of \ longitude$

<h3>Time difference</h3>

The time difference between the local apparent solar time and the Greenwich time is calculated as follows;

$t = 12 pm \ - \ 11 am\\\\t = 1 \ hour$

Since it is one hour time difference, the longitude is 15 degrees.

Learn more about Earth longitude here: brainly.com/question/1939015

5 0

1 year 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.