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lys-0071 [83]
3 years ago
11

A fish swims 12.0 m in 5.0 s. It swims the first 4.0 m in 2.0 s, the next 3.0 m in 1.2 s, and the last 5.0 m in 1.8 s. What is t

he average velocity of the fish during the time interval between 0.0 s and 3.2 s?
Physics
2 answers:
xz_007 [3.2K]3 years ago
7 0
Use the distance swan and the time elapsed in that interval.

Average velocity = distance / time

Average velocity = [4.0 m + 3.0m] / 3.2 s = 2.1875 m/s 
frosja888 [35]3 years ago
6 0

Answer:

IT IS OPTION D

Explanation:

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Answer:

increases.

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A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p
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Answer:

a) v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s}  =3.2051 \frac{m}{s}

b) v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s}  =5.2083 \frac{m}{s}

c) a=\frac{v_{2}-v_{1}  }{t_{2}-t_{1}  } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }

Explanation:

<em><u>The knowable variables are </u></em>

d_{t}=25m

t_{1}=1.3 s

t_{2}=3.9 s

t_{3}=5.5 s

Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is \frac{25}{3} m</u>

a) v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s}  =3.2051 \frac{m}{s}

b) v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s}  =5.2083 \frac{m}{s}

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) a=\frac{v_{2}-v_{1}  }{t_{2}-t_{1}  } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }

6 0
3 years ago
If the satellite was placed in an orbit three times farther away, about how long would it take to orbit the Earth once? Answer i
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Kepler's third law  is called the law of harmonies<span> which calculates the period and radius of orbit of a planet with the given dimensions and period of another planet. The comparison is proportional to the square of period and inversely proportional to the cube of the distance. Hence when the distance is placed three times the original ((3D)^3), the period should increase by sqrt of 27 times of 5.20 times the original period,</span>
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Read 2 more answers
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres o
SVETLANKA909090 [29]

Answer:

F = 7.7*10^{-10}N

Explanation:

You need to be careful with units for this problem. The force will be:

F =\frac{K*m1*m2}{d^2}

F=\frac{6.67259 * 10^{-11}*1.56*21.1*10^{-3}}{(5.34*10^{-2})^2}

F=7.7*10^{-10}N

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