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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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The value of g is greater at the poles than at the equator why

babunello [35]

Answer:

because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

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3 years ago

A police car chases a speeder along a straight road towards a cliff both vehicles move at 160km/h the siren on the police car pr

natta225 [31]

Answer:

f ’= 97.0 Hz

Explanation:

This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer

in this case the source is the police cases that go to vs = 160 km / h

and the observer is vo = 120 km / h

the relationship of the doppler effect is

f ’= f₀ (v + v₀ / v- $v_{s}$ )

let's reduce the magnitude to the SI system

v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s

v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s

we substitute in the equation of the Doppler effect

f ‘= 100 (330+ 33.33 / 330-44.44)

f ’= 97.0 Hz

4 0

3 years ago

What is the efficiency of the machine if the input is 263 J and the output is 142 J?

Papessa [141]

Answer: 57.79%

Explanation: 152J/263J=.577946768 or 57.79% or roundedthe nearest whole percent is 58%

7 0

3 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

3 years ago

A ball has an electric charge of +1.5 × 10-9 coulombs. At what distance from the ball's center is the electric field strength eq

Veronika [31]

F = kq1q2/r2
Where,
F - Coulomb Force
k - constant value which is equal to 8.98 × 10^9 newton square metre per square coulomb
q1 and q2 - two electric charges
r - distance.

5.8 * 10^5 = 1.5 * 10^-9 / r^2
5.8 * 10^5 r^2 = 1.5*10^-9
r^2 = 0.0000258620
r = 0.0050854694

So the distance is equal to 5.09 x 10^-3

3 0

3 years ago

Describe how one plays Dr.Dogeball​

Describe how one plays Dr.Dogeball