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bulgar [2K]
3 years ago
10

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 8.7 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 3.7 m/s. What upward force must a branch provide to support the swinging gibbon?
Physics
1 answer:
vlada-n [284]3 years ago
7 0

Answer:

198.505  N

Explanation:

Since this can be modeled as a point mass swinging system with the radius being the arm length. So r = 0.6m.

Also at it's lowest point the ape has a horizontal velocity v = 3.7 m/s. This velocity could result in a centripetal acceleration a, creating tension on the branch.

a = \frac{v^2}{r}

According to Newton's 2nd law, F = ma. So the tension is:

F = ma = \frac{mv^2}{r} = \frac{8.7*3.7^2}{0.6} = 198.5N

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A concrete highway is built of slabs 12 m long (20°C). How wide should the expansion cracks between the slabs be (at 20°C) to pr
Bezzdna [24]

Answer:

0.001152m

Explanation:

Linear expansivity of a material is the change in length of the material per unit length per degree rise in temperature. Mathematically,

¢ = ∆L/L1∆°C

¢ is the linear expansivity of the material = 12 x 10⁻⁶ °C⁻¹

Where ∆L is the change in length = L2-L1

L2 is the final length = ?

L1 is the initial length = 12m

∆°C is the change in temperature = °C2 - °C1 = 50-(-30) = 80°C

Substituting this values inside the formula to get the final length L2 after expansion, we have;

12 x 10⁻⁶ °C⁻¹ = L2-12/12×80

12 x 10⁻⁶ °C⁻¹ = L2-12/960

L2-12= 960×12 x 10⁻⁶ °C⁻¹

L2-12 = 0.001152

L2 = 12+0.001152

L2 = 12.001152m

Expansion will be the change in length L2-L1 = 12.001152-12

= 0.001152m

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5 0
3 years ago
Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assu
baherus [9]

Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

v = √2gh

h = 9 m and g = 9.8 m/s²

v = √(2 × 9 × 9.8) m/s

v = √176.4 m/s

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So his velocity just as he reaches the surface of the water is 13.28 m/s.

Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.

So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.

His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

J = (62 × 0 + 62 × 13.28) kgm/s = 0 +  823.46 kgm/s = 823.46 kgm/s

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6 0
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A c and e are correct but b and d are incorrect. i hope this helped
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