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MAXImum [283]
4 years ago
14

Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan

et increases linearly with the height of the atmosphere as measured from the top of a visible boundary layer, defined as 0 kilometers in altitude. The instruments on board can withstand a temperature of 601 K. At what altitude will the probe's instruments fail?

Physics
1 answer:
ZanzabumX [31]4 years ago
5 0
From my research, the image supports the question. From the graph given, we can construct the equation of the line using the two-point formula. Using the given value of 601 K, we can solve for the missing value of altitude.

y - y1 = [(y2 - y1)/(x2 - x1)](x- x1)
y - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4)

Substituting y = 601 to solve for x:
601 - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4<span>)
</span>x = 83

Therefore, the probe's instruments will fail at 83 kilometers.

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a capacitor is connected accross a bettery? why does each plate recieve a charge of equal to magnitude?
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A sharpening stone, also called whetstone, is used for sharpening ferrous tools. A round whetstone is mounted on the shaft of an
iVinArrow [24]

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3.2 kJ

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Since torque, τ = Iα where I = moment of inertia of round whetstone = 4.0 kgm² and α = angular acceleration of round whetstone.

So, α = τ/I

Given that τ = 20 Nm,

α = τ/I

α = 20 Nm/4.0 kgm²

α = 5 rad/s²

Using the equation for rotational motion, the angular velocity, ω = ω₀ + αt

where ω₀ = initial angular speed of round whetstone = 0 rad/s (since it starts from rest), ω = final angular speed of round whetstone, α = angular acceleration of round whetstone = 5 rad/s² and t = time of rotation = 8 s.

So, substituting the values of the variables into the equation, we have

ω = ω₀ + αt

ω = 0 rad/s + 5 rad/s² × 8s

ω = 0 rad/s + 40 rad/s

ω = 40 rad/s

So, its kinetic energy change ΔK = K₂ - K₁ = 1/2Iω² - 1/2Iω₀²

where K₁ = initial kinetic energy of round whetstone = 0 J (since the stone starts from rest) and K₂ = final kinetic energy of round whetstone after 8 s

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

K₂ - K₁ = 1/2Iω² - 1/2Iω₀²

K₂ - 0 J = 1/2 × 4.0 kgm² × (40 rad/s)² - 1/2 × 4.0 kgm² × (0 rad/s)²

K₂ = 2.0 kgm² × 1600 rad²/s² - 0 J

K₂ = 3200 kgm²rad²/s² - 0 J

K₂ = 3200 J

K₂ = 3.2 kJ

3 0
3 years ago
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