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3 years ago

12

Which of the following statements is true?

2 answers:

erica [24]3 years ago

6 0

An object at rest has an instantaneous acceleration of zero is true.

Option B

<u>Explanation</u>:

As we all know, acceleration is given by a, and formula is F/m, where F is force applied and m is mass of an object. Considering Instaneneous acceleration, it means the force per unit mass body is feeling during infinitesimally small time. Let us say 1 microsecond, acceleration of the body during that span of time is known as instanteneous acceleration. Considering option A, it might not be always true, as consider a vehicle going in straight direction, speed is changing and not direction. Option D is not again true, as consider earth rotating around sun, direction is changing but not speed. However, as explained above, instantaneous acceleration is always constant, and hence, body at rest have instantaneous acceleration zero.

rosijanka [135]3 years ago

3 0

The correct answers would be B, and d

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Answer:

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Explanation:

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Where $N$ is the number of turns

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$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

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$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

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For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

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