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Pavel [41]
3 years ago
8

Is the acceleration due to gravity more or less atop mt. everest than at sea level? defend your answer?

Physics
1 answer:
Nataly [62]3 years ago
3 0
The acceleration due to gravity is less at the top of mt. everest because its so far from the center of the earth .
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Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1320 K to a cold reservoir at 600 K. Calculate the
Zinaida [17]

Answer:0.0909 kJ/K

Explanation:

Given

Temperature of hot Reservoir T_h=1320 K

Temperature of cold Reservoir T_l=600 K

Heat of 100 kJ is transferred form hot reservoir to cold reservoir

Hot Reservoir is Rejecting heat therefore Q_1=-100 kJ

Heat is added to Reservoir therefore Q_2=100 kJ

Entropy change for system

\Delta s=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}

\Delta s=\frac{-100}{1320}+\frac{100}{600}

\Delta s=-0.0757+0.1666=0.0909 kJ/K

As entropy change is Positive therefore entropy Principle is satisfied

         

4 0
3 years ago
How do I solve this​
Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0
3 years ago
A manganese atom is pictured below.
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Manganese   has  2 (two) electron  that   would  free  floating   and   able  to  form  a  metallic  bond.
  The    electronic  configuration  of  manganese  is  (Ar)  3d5 4s2.  The   two   electron  in  4s  orbital  are  the  valence    electron  which  can  freely  move  from  one  place  to  another.
5 0
3 years ago
Read 2 more answers
What would be a good scientific model to write about?
drek231 [11]
The human Skeleton ? Lol
7 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

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

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

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

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

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

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

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

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
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