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balandron [24]
3 years ago
15

Make a 29 day timeline. Along the timeline draw and label the phases of the Moon starting with a full moon, crescent, first quar

ter, gibbous and full. Finish the second half showing the waning moons back to the new moon.
PLEASE Include the drawing and your answer to the question above

PLEASE NO FAKE ANSWERS.
Physics
1 answer:
Yanka [14]3 years ago
5 0
Ummm.... this is a type of question that u have to do by yourself for class. Good luck ;)!!!!! Plz mark as brainliest
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How does the input distance of a third-class lever compare to the output distance​
Alexandra [31]

Answer:

A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force

4 0
2 years ago
An electron is projected vertically upward with a speed of 1.70 ✕ 106 m/s into a uniform magnetic field of 0.320 t that is direc
Serggg [28]
The electron's path in the magnetic field is a straight line when viewed from above.

In fact, the electron initially moves upward, while the magnetic field is directed horizontally. The electron experiences a force due to the magnetic field (the Lorentz force), whose direction is given by the right-hand rule:
- index finger --> initial direction of the electron (upward)
- middle finger --> direction of the magnetic field (horizontally, away from the observer)
- opposite direction to the thumb* --> direction of the force (horizontally, but perpendicular to the magnetic field, to the right)

This means that the Lorentz force makes the electron moving perpendicular to the magnetic field in the horizontal plane, and since the direction of the field is not changing, this force does not change its direction, so the electron moves in the same direction of the force in the horizontal plane (to the right), therefore following a straight line.

* the direction should be reversed because the charge is negative.
7 0
3 years ago
A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin
gogolik [260]

Answer:

a) 23 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f}  (1)

  • The initial momentum p₀, can be written as follows:

       p_{o} =  m_{1}  * v_{1o} + m_{2}* v_{2o} =   6.0 kg * 5.0 m/s + 2.0 kg * v_{2o}  (2)

  • The final momentum pf, can be written as follows:

        p_{f} = (m_{1} + m_{2} )* v_{f}  = 8.0 kg* (-2.0 m/s)  (3)

  • Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

       v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg}  = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s  (4)

  • This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.
6 0
3 years ago
Una bola de acero rueda y cae por el borde de una mesa desde 4ft por encima del piso. Si golpea el suelo a 5ft de la base de la
ycow [4]

Answer:

     \large\boxed{\large\boxed{10ft/s}}

Explanation:

The question, translated, is:

  • <em>A steel ball rolls and falls off the edge of a table from 4ft above the floor. If you hit the ground 5ft from the base of the table, what was your initial horizontal velocity?</em>

<em />

<h2>Solution</h2>

<em />

This is a projectile motion, for which, the equations that you will need are:

    V_x_0= V_x=x\cdot t

     y=y_0+Vy_o\cdot t-g\cdot t^2/2

<u />

<u>1. Calculate the time that it takes the ball to fall 4ft</u>

    0=4ft-g\cdot t^2/2\\\\t^2=2\times 4ft/( 32.174ft/s^2)=0.24865s^2\\\\t=0.4986s

<u />

<u>2. Calculate the horizontal velocity:</u>

     V_x_0= V_x=x\cdot t\\\\V_x_0=5ft/0.4986s=10.027ft/s\approx 10ft/s

3 0
3 years ago
Can someone please help me with science.
vichka [17]
1 is b because a runs 20 and b rus 102 is c
5 0
3 years ago
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