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d1i1m1o1n [39]
3 years ago
10

José and Laurel measured the length of a stick's shadow during the day. Without knowing the length of the stick, which of their

measurements would you ask José and Laurel to examine more closely? A. 50 cm at 6:00 a.m. B. 75 cm at 12:30 p.m. C. 42 cm at 5:45 p.m. D. 55 cm at 7:15 p.m. dont speak spanish answer with english plz???
Physics
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer:

Option B.

Explanation:

Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).

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A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

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Weight, W = Mass, m × The acceleration due to gravity, g

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The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

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The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
2 years ago
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4 0
3 years ago
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Answer:

a. Ssystem  > 40 J/K

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Water, oil and silver are poured into the vessel
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The height of the oil column above the water in the vessel is determined as 2 cm.

<h3>Pressure of the vessel</h3>

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