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3 years ago

15

a car travelling initially at a speed of 30 m/s slows down at the rate of -2 m/s² for a distance of 60 meters. the car's velocit

y after covering this distance is __ m/s?

1 answer:

lana66690 [7]3 years ago

3 0

Answer:

answer

Explanation:

this is the answer with annotation and law

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gives the speed v versus time t for a 0.500 kg object of radius 6.00 cm that rolls smoothly down a 30° ramp. The scale on the ve

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Answer:

Rotational inertia of the object is given as

$I = 7.2 \times 10^{-4} kg m^2$

Explanation:

As we know that the acceleration of the object on inclined plane is given as

$a = \frac{gsin\theta}{1 + k^2/R^2}$

now we know that velocity at any instant of time is given as

$v = at$

now we know that if the graph between velocity and time is given then the slope of the graph will be same as acceleration

so here we have

$\frac{gsin\theta}{1 + k^2/R^2} = slope$

now from the graph slope of the graph is given as

$slope = \frac{3.5 - 0}{1}$

$\frac{gsin\theta}{1 + k^2/R^2} = 3.5$

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4 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

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