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mart [117]
3 years ago
7

An astronaut is being tested in a centrifuge. The centrifuge has a radius of 11.0 m and, in starting, rotates according to θ = 0

.390t2, where t is in seconds and θ is in radians. When t = 3.70 s, what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?
Physics
1 answer:
maxonik [38]3 years ago
8 0

Answer:

ω = 2.9 rad/s, v = 31.7 m/s, α = 0.78 rad/s², a = 91,6 m/s²

Explanation:

I will assume the given equation reads as Ф = 0.39t².

The angular velocity ω is the time derivative of Ф.

ω = 0.78t

The linear velocity v is given by ωr, where r is the radius of the centrifuge.

v = 0.78t r

The tangential acceleration α is the time derivative of ω.

α = 0.78

The radial acceleration a is given by: a = ω²r

a = (0.78t)²r

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The volume of rain that fells in the field is simply given by the area of the field, which is

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h=5.0 mm

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7 0
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The average intensity of sunlight at the top of the earth's atmosphere in 1390 w/m2. what is the maximum energy that a 34-m x 46
slavikrds [6]
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7 0
3 years ago
A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potentia
Llana [10]

Answer:

42.3 MV

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diameter of the metal sphere is given as

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r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

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V = \frac{kQ}{r}

V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}

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5 0
4 years ago
Which feature of a longitudinal wave corresponds to a trough in a transverse wave?
Rama09 [41]
It’s the crest, the crest is the top part of the wave and the trough is the bottom so they correspond
3 0
3 years ago
Read 2 more answers
*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the to
denis23 [38]

Answer:

μ₁ = 0.1048

μ₂ = 0.1375

Explanation:

Using  static equation can find in both point the moment and the forces so:

∑ M = F *d  , ∑ F = 0

∑ M A = 0

N₁ * 3 - 200 * 9.81 * 1.5 = 0

N₁ = 981  

∑ M y = 0

N₂ + 300 * ³/₅ - 981 - 20 * 9.81 = 0

N₂ = 997.2 N

∑ M C = 0

F₁ * 1.75  - 300 * ⁴/₅  * 0.75 = 0

F₁ = 102.86

∑ M B = 0

300 * ⁴/₅ * 1 - F₂ * 1.75 = 0

F₂ = 137.14 N

The Force F1 and F2 related the coefficients of static friction

F₁ = μ₁ * N₁   ⇒  102.86 N = μ₁ * 981 ⇒ μ₁ = 0.1048

F₂= μ₂ * N₂  ⇒  137.14 N = μ₂ * 997  ⇒ μ₂ = 0.1375

8 0
4 years ago
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