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lutik1710 [3]
1 year ago
8

Gibbons move through the trees by swinging from successive handholds, as we have seen. To increase their speed, gibbons may brin

g their legs close to their bodies. How does this help them move more quickly?
Physics
1 answer:
Sedbober [7]1 year ago
5 0

By bring their legs close to their bodies, they are decreasing the length of the pendulum which help them move more quickly.

A pendulum is nothing but a body suspended from a fixed point so that it can swing back and forth under the influence of gravity.

Here in this case Gibbons are bringing their legs close to their bodies and reducing the length of the pendulum. Since as the length of the pendulum increases the speed of the movement will be reduced. By bringing their legs close to their bodies they are reducing the length and in turn their speed increase and they move quickly.

To learn more about pendulum, visit here

brainly.com/question/14759840

#SPJ4

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Answer:

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Explanation:

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3 years ago
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3 years ago
A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge
Norma-Jean [14]

Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b  = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c  = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance    .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c)    ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt       ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π×10^{-7}×0.024 / 2π(1.20×10^{-2}) × ln (0.024 + 0.012/0.012) × 120

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current = 4 ×10^{-7}× 1.09861 × 120

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so here current in loops is 52.73 μA

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3 years ago
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