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3 years ago

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A fish is able to jump vertically out of the water with a speed of 4.45 m/s. What is the speed of the fish as it passes a point

0.6 m above the water?
How much time, after leaving the water does it take for the fish to pass a point 0.6 m above the water while it is on its way down?

1 answer:

rjkz [21]3 years ago

8 0

Answer:

Explanation:

given

initial velocity u = 4.45m/s

Height = 0.6m

g = 9.8m/s²

Required

final velocity v

Using the equation of motion;

v² = u²-2gH (upward motion of the fish makes g to be negative)

v² = 4.45²-2(9.8)(0.6)

v² = 19.8025-11.76

v² = 8.0425

v = 2.84 m/s

Hence the speed of the fish as it passes a point 0.6 m above the water is 2.84m/s

To get the time, we will use the formula

v = u - gt

2.84 = 4.45 - 9.8t

2.84-4.45 = -9.8t

-1.61 = -9.8t

t = 1.61/9.8

t = 0.164secs

Hence the time taken is 0.164secs

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A wrench is placed at 30 cm in front of a diverging lens with a focal length of magnitude 10 cm. What is the magnification of th

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Answer:

0.25

Explanation:

Magnification = image distance/object distance

mag = v/u.................. Equation 1

Given: f = -10 cm ( diverging lens) u = 30 cm.

Where can calculate for the value of v using

1/f = 1/u+1/v

make v the subject of the equation

v = fv/(u-f)..................... Equation 2

Substitute into equation 2

v = -30(10)/(30+10)

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v = -7.5 cm.

substituting into equation 1,

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A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, c

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Answer:

The velocities of the skaters are $v_{1} = 3.280\,\frac{m}{s}$ and $v_{2} = 0.024\,\frac{m}{s}$ , respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

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Where:

$m_{1}$ - Mass of the first skater, in kilograms.

$m_{2}$ - Mass of the second skater, in kilograms.

$v_{1,o}$ - Initial velocity of the first skater, in meters per second.

$v_{1}$ - Final velocity of the first skater, in meters per second.

$v_{b}$ - Launch velocity of the meter, in meters per second.

$v_{2}$ - Final velocity of the second skater, in meters per second.

If we know that $m_{1} = 70\,kg$ , $m_{b} = 0.043\,kg$ , $v_{b} = 32\,\frac{m}{s}$ , $m_{2} = 58.5\,kg$ and $v_{1,o} = 3.30\,\frac{m}{s}$ , then the velocities of the two people after the snowball is exchanged is:

By (1):

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$

$m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}$

$v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}$

$v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)$

$v_{1} = 3.280\,\frac{m}{s}$

By (2):

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$

$v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}$

$v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}$

$v_{2} = 0.024\,\frac{m}{s}$

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Answer:

Explanation:

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