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maxonik [38]
2 years ago
10

I need help with this answer I believe it's a democracy​

Physics
1 answer:
kipiarov [429]2 years ago
8 0

Answer:

a. democracy

Explanation:

beacouse the government control of their members

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Ayuda porfa es de actividades fisicas
igor_vitrenko [27]

Translate to English please

3 0
2 years ago
A 100 kg person rides in an elevator moving at a constant speed of
Airida [17]

Answer:

Explanation:

All this information only applies to the person. There is an extra tension force if we are talking about the elevator, but we are not. Dont forget to apply the units

Acceleration means change in speed or velocity. The elevator is moving at a constant speed of 3 meters. You wont even know you are moving because there is no change in acceleration. It equals 0

The forces ONLY acting on the person would be the force of gravity pulling them down, and the normal force that the elevator is reciprocating from the person standing on it.

Force = mass x acceleration. You have 100 kg and you are accelerating at 0 m/s. The force is 0. Which makes sense because the force of gravity and the net force completely cancel each other out.

8 0
3 years ago
W=90 kg×4n/kg<br>Please halp me i need it ​
kiruha [24]

I'm guessing that this is a problem to find the weight of a 90kg mass on a planet where the acceleration of gravity is 4 m/s^2. (Much less gravity than Earth, a little more than Mars.)

Just do the multiplication, and you get

360 Newtons.

5 0
3 years ago
A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?
Varvara68 [4.7K]

Answer:

<h2>15.25 N</h2>

Explanation:

       A force of 55\text{ }N is acting on a wagon along the road. The wagon weights 7.5\text{ }kg. Acceleration of the wagon is given as 5.3\text{ }\frac{m}{s^{2}}.

       Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

       Gravitational Force and Normal Reaction cancel out each other.

       Net External Force = Mass of system/wagon \times Acceleration of wagon

       F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N

F_{friction} has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0
3 years ago
Two cars approach a street corner at right angles to each
irakobra [83]

Answer:

the angle is given by

Tan theta = 35/59 = 0.59

so theta = Tan ^-1 ( 0.59 )

theta = 30.54 deg.

4 0
2 years ago
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