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wel
3 years ago
13

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

et varies linearly with the height of the atmosphere. The delicate instruments on board completely fail at a height of 61.5 kilometers. What is the approximate temperature at this altitude?
A. 200 K

B. 250 K

C. 300 K

D. 400 K

E. 450 K

Physics
1 answer:
lianna [129]3 years ago
8 0

Answer: E. 450 K

Explanation:

It is given that the temperature of the gaseous planet is linearly related with height of the atmosphere. we can write this in the mathematical equation:

y = m x +c

where y is the temperature values, x is height, m is the slope and c is the y-intercept. we have been given two sets of value in the image, using which we can find the value of slope in y-intercept.

at x = 18.40 km, y = 147.54 K

⇒147.54 = 18.40 m + c      

⇒c = 147.54 - 18.40 m  ..(1)

at x =78.11 km, y = 567.00 K

⇒567.00 =78.11 m + c       ..(2)

Put equation 1 in 2 and solve:

⇒567.00 =78.11 m + 147.54 - 18.40 m

⇒419.46 = 59.71 m

⇒ m =419.46 ÷59.71 = 7.025 K/km

c = 147.54 - 18.40 × 7.025 = 18.28 K

At height, x = 61.5 km the approximate temperature is :

y = 7.025 K/km ×  61.5 km + 18.28 K = 450.3 K

Thus, the approximate temperature at altitude 61.5 km is 450 K.

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Help solve these two problems im having trouble trying to start these problems?​
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Explanation:

25.

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We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):

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6 0
3 years ago
A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,
NISA [10]

Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N      

Explanation:

We have given mass of the crate m = 49.6 kg

Acceleration due to gravity g=9.8m/sec^2

Coefficient of static friction \mu _s=0.584

Coefficient of kinetic friction \mu _k=0.399

(a) Static friction force is given by F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N

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3 0
2 years ago
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Temka [501]

Source localization in ocean acoustics is posed as a machine learning problem in which data-driven methods learn source ranges directly from observed acoustic data: True.

<h3>What is machine learning?</h3>

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6 0
2 years ago
What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when
Gre4nikov [31]

Answer:

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Explanation:

Given:

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Mass of water = 60.2 g

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Object

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Mass of water displaced = 3.3 g

So, volume in water = 3.3 cm³

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Density = 19.24

So,

Object ,must be gold.

7 0
2 years ago
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