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Sveta_85 [38]
4 years ago
8

Which of these physical settings makes sense for an object moving along the x-axis? Its position at any time is given by x(t) =

3cos(π t) + 2
a. A skydiver falling from a plane, before she opens her parachute.
b. The shadow cast by a rock that's stuck on the edge of a spinning wheel.
c. A car driving from San Diego to Los Angeles.
d. An ant walking at a steady pace from the tip of a propeller toward the hub.
e. None of these settings.
Physics
1 answer:
kow [346]4 years ago
4 0

Answer:

option b is correct.

Explanation:

Since the position as a function of time is given by

x(t)=3cos(\pi t)+2 which is a periodic function thus the motion of the particle repeats after a time of 2 seconds thus among the given options only option b is case of repetitive motion thus option b is most appropriate since the spinning wheel repeates it's motion.

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A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the ce
Dmitry [639]

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

6 0
3 years ago
The net flow of energy into and out of the earths system is referred to as energy budget. which type of energy is lost in space
Svet_ta [14]
D not 100% sure though.
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3 years ago
What would u expect the tides to be like during a first quarter moon?
kicyunya [14]

Answer:

During the first quarter or last quarter phase of the moon, when the sun and moon are perpendicular (at right angles) to each other in relation to the Earth, the tidal gravitational pulls interfere with each other, producing weaker tides, known as neap tides.

Explanation:

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3 years ago
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The main difference between baseline activities and health-enhancing activities is that baseline activities maintain the status quo, while health-enhancing activities make things better.
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One end of a string 6.26 m long is moved up and down with simple harmonic motion at a frequency of 95 Hz . The waves reach the o
uranmaximum [27]

Answer:

Explanation:

Given

length of string L=6.26\ m

frequency f=95\ Hz

time taken by wave to reach at other end t=0.5\ s

speed of  wave is given by

v=\frac{length\ of\ string}{time\ taken}

v=\frac{6.26}{0.5}

v=12.52\ m/s

wavelength of is given by

\lambda =\frac{velocity}{frequency}

\lambda =\frac{12.52}{95}

\lambda =0.131\ m                          

4 0
3 years ago
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