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pashok25 [27]
3 years ago
11

1) The astronaut's sense of imbalance in a space flight is due to the rotation of a device in which he or she feels a fictitious

gravity that mimics gravity. The the device rotates according to the mathematical equation: n =g^0.5/2πr^0.5 Where n is the rotational speed and is measured in rotations per second (s). (r) is the radius of the circulator and is measured in meters (m). (g) is a simulation of gravity. Calculate the rotation speed of a device, that has a radius of 1.7 meters and rotates to simulate the gravitational force which equals to: g = 9.8 m/s^2 with steps please
Mathematics
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

n = 0.38 rotation per second

Step-by-step explanation:

The device rotates according to the mathematical equation :

n=\dfrac{g^{0.5}}{2\pi r^{0.5}} ...(1)

Where n is the rotational speed and is measured in rotations per second (s).

r is radius of the circulator and is measured in meters (m).

We need to find the rotational speed when radius is 1.7 m. Put r = 1.7 m in equation (1).

n=\dfrac{(9.8)^{0.5}}{2\pi \times (1.7)^{0.5}}\\\\=0.38\ \text{rotations per second}

Hence, the rotational speed is 0.38 rotation per second.

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Which series of transformations will not map figure H onto itself
7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation (x+0,y-2) will map vertices of the square into points

  • (2,1)\rightarrow (2,-1);
  • (1,2)\rightarrow (1,0);
  • (2,3)\rightarrow (2,1);
  • (3,2)\rightarrow (3,0).

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

  • (2,-1)\rightarrow (2,3);
  • (1,0)\rightarrow (1,2);
  • (2,1)\rightarrow (2,1);
  • (3,0)\rightarrow (3,2)

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation (x+2,y-0) will map vertices of the square into points

  • (2,1)\rightarrow (4,1);
  • (1,2)\rightarrow (3,2);
  • (2,3)\rightarrow (4,3);
  • (3,2)\rightarrow (5,2).

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

  • (4,1)\rightarrow (2,1);
  • (3,2)\rightarrow (3,2);
  • (4,3)\rightarrow (2,3);
  • (5,2)\rightarrow (1,2)

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation (x+3,y+3) will map vertices of the square into points

  • (2,1)\rightarrow (5,4);
  • (1,2)\rightarrow (4,5);
  • (2,3)\rightarrow (5,6);
  • (3,2)\rightarrow (6,5).

2nd transformation = reflection over y = -x + 7 will map vertices into points

  • (5,4)\rightarrow (3,2);
  • (4,5)\rightarrow (2,3);
  • (5,6)\rightarrow (1,2);
  • (6,5)\rightarrow (2,1)

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation (x-3,y-3) will map vertices of the square into points

  • (2,1)\rightarrow (-1,-2);
  • (1,2)\rightarrow (-2,-1);
  • (2,3)\rightarrow (-1,0);
  • (3,2)\rightarrow (0,-1).

2nd transformation = reflection over y = -x + 2 will map vertices into points

  • (-1,-2)\rightarrow (4,3);
  • (-2,-1)\rightarrow (3,4);
  • (-1,0)\rightarrow (2,3);
  • (0,-1)\rightarrow (3,2)

These points are not the vertices of the initial square.

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3 years ago
Find x if x + y = 16, x * 3 is even, x is a factor of y, and 2x+y= 20.
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Answer:

x = 0

Step-by-step explanation:

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For his trip on the way there:
D = 7*s

For his trip on the way home you have:
D = 5*(s+18)

use substitution to solve
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