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antoniya [11.8K]

3 years ago

9

Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver springs upward with an initial speed o

f 2.22 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?

1 answer:

adell [148]3 years ago

7 0

Answer:

Part a)

$v_f = 7.99 m/s$

Part b)

$y = 3.25 m$

Explanation:

Part a)

Since the diver is moving under gravity

so here its acceleration due to gravity will be uniform throughout the motion

so here we will have

$v_f^2 - v_i^2 = 2 a y$

here we have

$v_i = 2.22 m/s$

$y = -3 m$

$v_f^2 - (2.22)^2 = 2(-9.81)(-3)$

$v_f = 7.99 m/s$

Part b)

at highest point of his motion the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a (\Delta y)$

$0 - 2.22^2 = 2(-9.81)(y - 3)$

$y = 3.25 m$

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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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Which statement describes the magnetic field inside a bar magnet? It points from north to south. It points from south to north.

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Magnetic field lines emerge from the North pole and end in the South pole of a bar magnet.

Inside the magnet are also present inside the bar magnet and never intersect at any point.

Therefore, inside the bar magnet, the magnetic field points from north to south.

To know more about Magnetic Field:

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