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Sedbober [7]
2 years ago
15

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.2 m/s .
What upward force must a branch provide to support the swinging gibbon?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
aleksklad [387]2 years ago
5 0

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force  T  is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force  in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T =  mg -  mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

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a conducting rod whose length is 25 cm is placed on a u-shaped metal wire that has a resistance of 8.0 ω . the wire and the rod
andreyandreev [35.5K]

Answer:

The current is I  =  1 A

The direction is anti-clockwise

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

     the length of the conducting rod is L  = 25 \ cm =  \frac{25}{100}  =  2.5 \ m

     The resistance is  R = 8  \Omega

      The magnetic field is  B = 0.40\ T

        The speed of the rod is v = 6.0 m/s

The emf induced is

          \epsilon  = BLv

 substituting values we have

           \epsilon = 0.40 * 2.5 * 8

           \epsilon = 8V

From ohm law the induced current would be

      I  =  \frac{\epsilon}{R }

 substituting values we have

       I  =  \frac{8}{8 }      

       I  =  1 A

The direction anticlockwise this because according to lenze law the current due to change in magnetic field will act in the opposite direction of the  force causing the magnetic field to change

4 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
2 years ago
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Explanation:

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This means that energy is inversely proportional to wavelength. So, more is the energy of an electromagnetic radiation less will be its wavelength.

Also,          f = \frac{c}{\lambda}

Hence, less will be the wavelength more will frequency of a radiation.

Gamma rays are the rays that have highest energy, small wavelength and highest frequency.

Thus, we can conclude that gamma rays are the electromagnetic radiation which has the highest frequency.

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<span>Object B stays neutral but becomes polarized.

Hope this helps.

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Why does a wave undergo a 180° phase change at a fixed boundary?
Leno4ka [110]

Answer:

The phase change of 180^{o} can be theoretically understood as follows:

For transmission or propagation of waves between media the wave motion should maintain a principle of continuity meaning that the wave function at the interface should be continuous and diffrentiable at the interface.

At the point of incidence there are 2 types of waves reflected wave and the incident wave. Now the principle of continuity dictates that the sum of the phases of the above 2 waves should be same as that of transmitted wave. If we use these relations we notice that the reflected wave shall either change it's phase by 180^{o} or will not change it's phase depending on the relationship between the refractive indices of the incident and the reflecting medium. For a solid boundary a phase change of 180^{o} occurs.

6 0
3 years ago
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