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2 years ago

PLEASE HELP ASAP.

Physics

1 answer:

kifflom [539]2 years ago

4 0

Volume of the original block (V) of length (l), width (w) and height (h) is given by the following equation

$V = l*b*h$

hence volume of original block, V = (6)(4)(10) = $240 cm^{3}$

Now say we cut this block in two equal parts, than the length of new block will be $6/2 = 3 cm$ ; where as its width and height will be same as of original block i.e. 4 cm and 10 cm respectively.

So, Volume of half-block is $= (3)(4)(10) = 120 cm^{3}$ ; or we can say that the volume of half-block is half that of orignal block.

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To understand the experiment that led to the discovery of the photoelectric effect.

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Answer:

A) Emin = eV

B) Vo = (E_light - Φ) ÷ e

Explanation:

Energy of electron is the product of electron charge and the applied potential difference.

The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;

Emin = eV

The maximum stopping potential energy is eVo,

The energy of the electron due to the light is E_light.

If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy

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Five different forces act on an object. Is it possible for the net force on the object to be zero?

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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.