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Bas_tet [7]
3 years ago
10

In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5220km away, and released. It found its way

back to its nest a time 12.8days after release.a. If we place the origin in the nest and extend the +x-axis to the release point, what was the bird's average velocity for the return flight?v= ___.b. What was the bird's average velocity in m/s for the whole episode, from leaving the nest to returning?v= ___.
Physics
1 answer:
denis-greek [22]3 years ago
3 0

Answer:

-4.72005 m/s

0 m/s

Explanation:

Displacement = 5220 km

Time = 12.8 days

The average velocity is given by

v_a=\dfrac{0-5220\times 10^3}{12.8\times 24\times 60\times 60}\\\Rightarrow v_a=-4.72005\ m/s

The average  for the return flight is -4.72005 m/s

When the total displacement is divided by the total time of a journey then we get the average velocity.

Displacement is the minimum distance between the initial and final points of the journey.

Here, the displacement of the whole episode is 0 as the initial and final point is zero.

Hence, the average velocity for the whole episode is 0 m/s

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nignag [31]

The decimal point is placed after two digits starting from the end. For each decimal place, we can write the number divided by 100.

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Divide the numerator and denominator by 2:

\frac{2112}{100}= \frac{156}{50}

The numerator and denominator can be divided by 2 again:

\frac{78}{25}

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Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
Karen has a mass of 51.9 kg as she rides the up escalator at Woodley Park Station of the Washington D.C. Metro. Karen rode a dis
kifflom [539]

Answer:

28852 J

Explanation:

When a force applied in a body produces a displacement in it, the force realized a work. The force that moves Karen is contrary to her weight and must be equal to it.

The work (W) is:

W = F.d.cos(θ), where F is the force, d is the displacement, and θ is the angle.

Knowing that cos(26°) = 0.899, and F = m*g

W = 51.9*9.8*63.1*0.899

W = 28852 J

4 0
3 years ago
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