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3 years ago

Why do we never see the moon next to, say, polaris?.

1 answer:

6 0

Polaris is near the North Celestial Pole and nowhere near the Ecliptic. The Moon's orbit is tipped by 5 degrees to the Ecliptic. An eclipse can only happen when the Moon is near the line of nodes.

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Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec

Wittaler [7]

Answer:

The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value intro the formula

$300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v$

$v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}$

$v=-0.5\ m/s$

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2$

$K.E_{i}=0.31875\ J$

We need to calculate the total final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2$

$K.E_{f}=0.1125\ J$

We need to calculate the energy lost during the collision

Using formula of energy lost

$energy\ lost=\dfrac{0.31875-0.1125}{0.31875}$

$energy\ lost=\dfrac{11}{17}\ J$

Hence, The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

3 0

3 years ago

The metric unit of force is the _____. A.newton B.pound C.kilogram D.watt

hammer [34]

Force = mass × acceleration = kg × m/s^2 = Newton

3 0

4 years ago

Read 2 more answers

Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,

adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

x_cm = 1 /M ∑ $x_{i}$ m_{i}

y_cm = 1 /M ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

x4=?

y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

M = 2.00 + 2.20 + 3.40

M = 7.6 kg

0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

x4 = -6.8 / 6

x4 = -1,133 m

Axis y

0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

y4 = -11/6

y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0

4 years ago

Help a homie plz

Sever21 [200]

Answer:

the 4 resistor wired in parallel is equal to 23.946ohm

4 0

3 years ago

Briefly describe the history of the metric system as it applies to the meter and how the definition of a meter has changed over

Shalnov [3]

Answer and explanation;

In 1670 Gabriel Mouton, Vicar of St. Paul’s Church and an astronomer proposed the swing length of a pendulum with a frequency of one beat per second as the unit of length.

In 1791 the Commission of the French Academy of Sciences proposed the name meter to the unit of length. It would equal one tens-millionth of the distance from the North Pole to the equator along the meridian through Paris.It is realistically represented by the distance between two marks on an iron bar kept in Paris.

In 1889 the 1st General Conference on Weights and Measures define the meter as the distance between two lines on a standard bar that made of an alloy of 90%platinum with 10%iridium.

In 1960 the meter was redefined as 1650763.73 wavelengths of orange-red light, in a vacuum, produced by burning the element krypton (Kr-86).

In 1984 the Geneva Conference on Weights and Measures has defined the meter as the distance light travels, in a vacuum, in 1299792458⁄ seconds with time measured by a cesium-133 atomic clock which emits pulses of radiation at very rapid, regular intervals.

8 0

4 years ago