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inysia [295]
3 years ago
15

An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205

m). By how much (in newtons) does her weight change when she goes from sea level to the observatory?
Engineering
1 answer:
lara [203]3 years ago
8 0

Answer:

0.845\ \text{N}

Explanation:

g = Acceleration due to gravity at sea level = 9.81\ \text{m/s}^2

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2

Weight at sea level

W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}

Weight at the given height

W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}

Change in weight W_h-W=636.805-637.65=-0.845\ \text{N}

Her weight reduces by 0.845\ \text{N}.

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Drivers must be careful when driving close to cyclists and should keep at least ___ feet apart when passing cyclists on the road
faust18 [17]

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at least 8 feet

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2 years ago
Water (cp = 4180 J/kg·°C) enters the 2.5 cm internal diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rat
aksik [14]

Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

Diameter, D = 2.5 cm

Temperature of water in =  17°C

Temperature of water out = 80°C

mass rate of water =1.8 kg/s.

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U = 700 W/m2 ·°C

Since Temperature of steam is at saturation,

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Find attached the full solution to the question.

3 0
3 years ago
typedef struct bitNode { int data; struct bitNode *left; struct bstNode *right; } bstNode; int solve(bstNode* root) { if (root =
Sergio039 [100]

Answer:

The ten numbers to be filled in the blanks are: 18, 7, 7, 11, 18, 36, 3, 8, 13, 50.

Explanation:

keeps on going to the left node until node->left == NULL;

now at Node 18;

left = right = 0; hence condition is not satisfied

18 is printed first.

the value 18 is returned .

Then we reach at 4;

from there we move to 7;

just like 18, similar things happen with 7 and 7 is printed, the value 7 is returned.

Now coming to Node 4,

left = 0, right = 7 ; hence the condition is satisfied & res = 7; 7 is printed.

For Node 16, left = 7 ; right = 11(but for this we visit 11 first and 11 is printed)

for 16; condition is satisfied; res = 7 + 11 = 18 ; 18 is printed

Now for 5; left = right = 18; the condition is satisfied; so res = 18 + 18 = 36; 36 is printed

Next we visit Node 3; 3 is printed & 3 is returned

Then Node 8 ; 8 is printed & 8 is returned

for Node 13; left = 3, right = 8 ; condition is not satisfied, 13 is printed.

For Node 50; left = 36 right = 13 ; condition is not satisfied hence 50 is printed.

So the order of printing is  18 7 7 11 18 36 3 8 13 50.

4 0
3 years ago
A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled
Vikentia [17]

Answer:

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The spacing between each wall and the plate, d = 10 mm = 0.01 m

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Suppose the drag force that exist between each wall and plate is F and F' respectively:

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A = Cross - sectional Area

Therefore,

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Pressure, P = \frac{F}{A}

Therefore,

\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}

\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}

8 0
2 years ago
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