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2 years ago

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 8.6 kg gibbon

has an arm length (hand to shoulder) of 0.60 mm. We can model its motion as that of a point mass swinging at the end of a 0.60-mm-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.5 m/s.
Required:
What upward force must a branch provide to support the swinging gibbon?

1 answer:

fiasKO [112]2 years ago

5 0

Answer:

Explanation:

i dont knoe sorry

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Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the hear

olchik [2.2K]

Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V

So minimum capacitance will be determined from;

i(t) = C(dv/dt)

So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]

C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF

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2 years ago

An object moving at a constant velocity travels 274 m in 23 s. what is its velocity?

svetlana [45]

V= 274 meters / 23 sec

V= 11.91 meters per sec

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2 years ago

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An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the

tamaranim1 [39]

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

Attached is the full solution

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2 years ago

A slender rod is 80.0 cm long and has mass 0.370 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

nataly862011 [7]

Answer:

1.10 m/s

Explanation:

Linear speed is given by

$v=r\omega$

Kinetic energy is given by

$KE=0.5I\omega^{2}$

Potential energy

PE= mgh

From the law of conservation of energy, KE=PE hence

$0.5I\omega^{2}=mgh$ where m is mass, I is moment of inertia, $\omega$ is angular velocity, g is acceleration due to gravity and h is height

Substituting m2-m1 for m and 0.5l for h, $\frac {2v}{L}$ for $\omega$ we obtain

$0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)$

$(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I}$ and making v the subject

$v^{2}=\frac {gl^{3}(m2-m1)}{4I}$

$v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}$

For the rod, moment of inertia $I=\frac {ML^{2}}{12}$ and for sphere $I=MR^{2}$ hence substituting 0.5L for R then $I=M(0.5L)^{2}$

For the sphere on the left hand side, moment of inertia I

$I=m1(0.5L)^{2}$ while for the sphere on right hand side, $I=m2(0.5L)^{2}$

The total moment of inertia is therefore given by adding

$I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}$

Substituting $\frac {L^{2}(M+3m1+3m2)}{12}$ for I in the equation $v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}$

Then we obtain

$v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}$

This is the expression of linear speed. Substituting values given we get

$v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s$

8 0

2 years ago

It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica

denis23 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is $\theta = 24.2^o$

Explanation:

From the question we are told that

The length of eire let out is $L = 250 \ m$

The emf generated is $\epsilon = 40 V$

The earth magnetic field is $B = 5.0 *10^{-5} T$

The speed of the shuttle and tether is $v = 7.80 * 10^3 \ m/s$

The emf generated is mathematically represented as

$\epsilon = L\ v\ B\ sin \ \theta$

making $\theta$ the subject of the formula

$\theta = sin ^{-1}[ \frac{\epsilon}{L * B *v} ]$

substituting values

$\theta = sin ^{-1}[ \frac{40}{250 * (5*10^{-5}) *(7.80 *10^{3})} ]$

$\theta = 24.2^o$

6 0

2 years ago