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4 years ago

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A scalloped hammerhead shark swims at a steady speed of 1.5 m/s with its 88 cm -cm-wide head perpendicular to the earth's 56 μT

magnetic field. What is the magnitude of the emf induced between the two sides of the shark's head? Express your answer using two significant figures.

1 answer:

Nadya [2.5K]4 years ago

8 0

Answer:

7.4 x 10⁻⁵ T.

Explanation:

Emf induced = BLv

Where B is magnetic field , L is length of rod or any other object moving in magnetic field and v is velocity of moving rod.

Here B = 56 x 10⁻⁶ T

L = 88 X 10⁻² m

v = 1.5 m/s

emf induced = 56 x 10⁻⁶ x 88 x 10⁻² x 1.5

= 7.4 x 10⁻⁵ T.

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The work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

$F= \frac{mv}{t}$

The force applied to an object increases with increases in the velocity of the object.

In the given diagram, the resultant velocity of the puck is calculated as follows;

Figure a:

$\Delta v = v_f -v_i\\\\\Delta v = 5 - 6 = - 1 \ m/s$

Figure b:

$v = \sqrt{4^2 + 3^2} \\\\v = 5 \ m/s$

Figure c:

$\Delta v = 4 - (-2)\\\\\Delta v = 6 \ m/s$

Thus, the work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

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3 years ago

8. In the billiard ball room at the basement of Memorial Union, two identical balls are traveling toward each other along a stra

mel-nik [20]

Answer:

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s

Explanation:

In an elastic collision, both the total momentum and the total kinetic energy of the system are conserved.

The conservation of the momentum can be written as:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where

$m_1$ is the mass of ball 1

$m_2$ is the mass of ball 2

$u_1$ is the initial velocity of ball 1

$u_2$ is the initial velocity of ball 2

$v_1$ is the final velocity of ball 1

$v_2$ is the final velocity of ball 2

The conservation of kinetic energy can be written as

$\frac{1}{2}m_1 u_2^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Working together the two equations, it is possible to find two expressions for the final velocities in terms of the initial velocities:

$v_1=\frac{m_1 -m_2}{m_1 +m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\v_2=\frac{2m_1}{m_1+m_2}u_1 -\frac{m_1 -m_2}{m_1 +m_2}u_2$

In this problem we have:

$m_1 = m_2 = m$ since the mass of the two balls is identical

$u_1=+4.67 m/s$ is the initial velocity of ball 1

$u_2=-7.89 m/s$ is the initial velocity of ball 2

Substituting into the equations, we find the final velocities:

$v_1=\frac{2m}{m+m}u_2=u_2 =-7.89 m/s\\v_2=\frac{2m}{m+m}u_1=u_1=+4.67 m/s$

Therefore:

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s

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3 years ago

A rectangular block of copper metal has a mass of 100 g. The dimensions of the block are 2 cm by 2 cm by 2 cm. From this data, w

lina2011 [118]

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3 years ago

Read 2 more answers

A runner starts from rest and in 2 s reaches a speed of 7 m/s. If we assume that the speed changed at a constant rate (constant

PtichkaEL [24]

Answer:

$v=3.5\frac{m}{s}$

Explanation:

The average speed is defined as:

$v=\frac{\Delta x}{\Delta t}$

Using the equations for uniformly accelerated motion, we calculate the runner's acceleration:

$a=\frac{v_f-v_o}{2}\\a=\frac{7\frac{m}{s}-0\frac{m}{s}}{2s}\\a=3.5\frac{m}{s^2}$

Now, we can calculate the distance that the runner travels:

$\Delta x=v_0t+\frac{at^2}{2}\\\Delta x=(0\frac{m}{s})(2s)+\frac{3.5\frac{m}{s}(2s)^2}{2}\\\Delta x=7m$

Finally, we calculate the runner's average speed:

$v=\frac{7m}{2s}\\v=3.5\frac{m}{s}$

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3 years ago

It takes a photon 8 minutes and 25 seconds ro reach earth. What is the distance (labeled 1 au) in meter between the sun and eart

tangare [24]

Answer:

$x=0.017 AU$

Explanation:

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$x=0.017 AU$

I hope it helps you!

3 0

3 years ago