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FromTheMoon [43]
3 years ago
10

Scientists are studying a moving glacier. To monitor the flow of the glacier, they place a series of five markers, A, B, C, D, a

nd E, in a straight line across the path of flow. A and E are closest to the edges of the glacier. C is in the center of the glacier. B and D are between A and C and C and E respectively. How would the scientists predict the motion of each of the markers relative to the edges of the valley down which the glacier flows? What pattern would they predict in the markers over time?
Physics
1 answer:
Blababa [14]3 years ago
4 0

Answer: c a d b

Explanation:

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The following problem presents hypothetical values. In the diagram below, the synodic period (one lunar month) is given as 30 da
balu736 [363]

Answer:

x is approximately 95.495

Explanation:

The given information are;

One lunar month = 30 days

The measured time between the first and third quarters = 14.9 days

Therefore, the angle of rotation for 30 days = 360°

The angle of rotation for 14.9 days = 14.9×360/30 = 178.8°

Therefore, whereby 2·θ = 178.8°

θ = 178.8°/2 = 89.4°

Therefore, given that the angle at the tangent from Sun to the Moon = 90°, we have;

The angle between the line from the Sun to the Moon = 90° - 89.4° = 0.6°

From sine rule, we have;

1/(sin 0.6) = x/(sin 90°)

x = (sin 90°)/(sin 0.6°) ≈ 95.495

x = 95.495.

6 0
3 years ago
PLEASE I NEED ANSWER BY 7:40!!!!!!!
Mashutka [201]

Answer:

Student A is correct

Explanation:

Colored objects look the way they do because of reflected light. When sunlight is shined on a green leaf, the violet, red and orange wavelengths are absorbed. The reflected wavelengths appear green. In each case we are seeing the complementary colors to the ones absorbed.

6 0
3 years ago
A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ
valentinak56 [21]

Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

or,                   v = \sqrt{\frac{2K.E}{m}}

                        = \sqrt{2 \times 3.6 J}{0.2 kg}

                        = \sqrt{36}

                        = 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0
3 years ago
The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
Is telekinesis real??? I really wanna start learning it!
Lesechka [4]

Answer:

if you only have to control your chakra and know how to get all your vibes to pass it to objects and it takes time to practice

4 0
3 years ago
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