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Stolb23 [73]
3 years ago
12

If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?

Physics
1 answer:
aivan3 [116]3 years ago
4 0

Initial velocity = \(v_0\)

acceleration in the downward direction = -9.8 \(\frac {m}{s^2}\)

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

\(v^2 = {v_0}^{2} + 2aH

\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410

\({v_0}^{2} = 2 \times 9.8 \times 0.410\)

\(v_0 = 2.834 \frac {m}{s}\)

Speed with which the flea jumps = \(2.834 \frac {m}{s}\)

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Answer:

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

Explanation:

The average emf induced can be calculated from the formula

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

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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

Magnetic flux is given by the formula

\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}

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A = 0.0225\pi

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

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

( NOTE: cos (0^{o}) = 1 )

\phi_{i}  = 0.03375\pi Wb

For \phi_{f}

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

Then

\phi_{f} = 0

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\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

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

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

Emf = 37.9 V

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A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it
mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

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v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

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Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

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With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

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Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

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y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

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Answer:

Thank you so much! Have a great day!

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