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dsp73
3 years ago
7

Describe how one plays Dr.Dogeball​

Physics
1 answer:
JulsSmile [24]3 years ago
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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Balance the equation-<br> Al+Mn02 ———-&gt; Mn + Al2O3
joja [24]

Answer:

                                                                                                         

Explanation:

3 0
3 years ago
Newer telephone circuits, built during the last decade, offer higher quality because they were built using analog transmission.
hammer [34]

Answer:

Well, newer telephone circuits built during the last decade are based on the digital transmission, not on the analog transmission. So it's the digital transmission circuit that has made the higher quality. Digital circuits converts the voice signals into the binary codes which is then translated again into the voice signal at the receiving end.

The answer is false.

Explanation:

6 0
3 years ago
8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-
Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula</u>:  

A=A_{o}.2^{\frac{-t}{H}} (1)  

Where:  

A=0.375 g is the final amount of the material  

A_{o}=3 g is the initial amount of the material  

t=1 h is the time elapsed  

H is the half life of the material (the quantity we are asked to find)  

Knowing this, let's substitute the values and find h from (1):

0.375 g=(3 g)2^{\frac{-1h}{H}} (2)  

\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}} (3)  

Applying natural logarithm in both sides:

ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}}) (4)  

-2.079=-\frac{1 h}{H}ln(2) (5)  

Clearing H:

H=\frac{-1h}{-2.079}(0.693) (6)  

Finally:

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4 0
3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

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b )

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c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
What is impulse? How does this relate to momentum?
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4 0
3 years ago
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