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3 years ago

Describe how one plays Dr.Dogeball

1 answer:

4 0

•To play Dr. Dodgeball you need to have 2 teams to verse each other.
•Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors)
•Continue to play dodgeball how you normally would
•When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge.
•Finally, try to get all the doctors and players out from the other team. Get the doctors first, for they cannot revive themselves. Which means the other players are out after they get hit with a ball since the doctors are out. (Some games are played where if all doctors are out the game ends)
Hope this helped! Play on! And plz mark brainliest lol this was long to write :D

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3 years ago

Three beads are placed along a thin rod. The first bead, of mass m1 = 28 g, is placed a distance d1 = 1.5 cm from the left end o

Vladimir79 [104]

Answer:

Part a)

Center of mass with respect to the left end is given as

$r_{cm} = 5.63 cm$

Part b)

Center of mass with respect to middle bead is

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

Part c)

Center of mass with respect to middle bead is

$r_{cm} = 1.63 cm$

Explanation:

Part a)

As we know that the center of mass of the system of mass is given by the formula

$r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

here we have

$m_1 = 28 g$

$m_2 = 11 g$

$m_3 = 45 g$

$r_1 = 1.5 cm$

$r_2 = 1.5 + 2.5 = 4 cm$

$r_3 = 1.5 + 2.5 + 4.6 = 8.6 cm$

Now we have

$r_{cm} = \frac{28(1.5) + 11(4) + 45(8.6)}{28 + 11 + 45}$

$r_{cm} = 5.63 cm$

Part b)

As we know that the center of mass of the system of mass is given by the formula

$r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

here we have

$m_1 = 28 g$

$m_2 = 11 g$

$m_3 = 45 g$

$r_1 = -d_2 = -2.5cm$

$r_2 = 0$

$r_3 = d_3 = 4.6 cm$

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

Part c)

Now plug in the values in above formula

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

$r_{cm} = \frac{28(-2.5) + m_2(0) + 45(4.6)}{28 + 11 + 45}$

$r_{cm} = 1.63 cm$

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3 years ago

How does object float

IceJOKER [234]

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3 years ago

Read 2 more answers

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Lina20 [59]

Answer:4.8 m/s^2

Explanation:

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Force=50.5N

Acceleration =force ➗ mass

Acceleration =50.5 ➗ 10.5

Acceleration =4.8 m/s^2

8 0

4 years ago

Describe how one plays Dr.Dogeball​

Describe how one plays Dr.Dogeball