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3 years ago

True or False: In a parallel circuit, if one path is broken, electrons continue to flow to the

Physics

1 answer:

Snezhnost [94]3 years ago

4 0

Answer:

false

Explanation:

this is what google said

No because the path the electricity needs to follow is broken. In a parallel circuit, electricity has more than one path to follow. Electrons can follow different paths as they flow from the negative side of the battery to the positive side.

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Two common ions of manganese are Mn2+ and Mn3+. Which binary ionic compound is likely to form? A) Mn2N3 B) MnBr4 C) MnCl D) MnO

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Given that Oxygen has an oxidation state of 2 -, you can combine Mn 3+ with O 2- ions to form Mn2O3, and you can combine Mn 2+ with O 2- to form Mn2O2 which is MnO.

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3 years ago

What is the momentum of a 50-kilogram ice skater gliding across the ice at a speed of 5 m/s? (1 point)

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Momentum is a term used to quantify the motion of an object has. It is calculated as the the product of the object's mass and the velocity. It is expressed as:

Momentum = m x v
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Therefore, the correct answer is the last option.

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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.