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zhenek [66]
2 years ago
7

Which element is likely more reactive, and why?

Physics
1 answer:
erica [24]2 years ago
6 0
Element 2 explanation just trust
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Differences between looping and simersaulting​
Anika [276]
Somersaulting- for longer distances.It bends the narrow end in the direction it wants to go & takes grip with tentacles. It releases the broad end and straightens up. like this it continues. looping- for shorter distances.


Hope this helps
3 0
3 years ago
two children having a disagreement pull on a 50N sled in opposite directions. one pulls with a force of 300N east, the other wit
prohojiy [21]
<span>Since forces are vector quantities, we must indicate direction using positive and negative values. East will be assigned positive and west will be negative. Friction will act as a negative force since it impedes action. To calculate the net force we sum the vector quantities, as follows. Net force equals 50n which is derived by the following calculation: 300n-220n-30n.</span>
7 0
3 years ago
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Given the distance between the crest of one wave and the crest of the next wave, you can determine the?
Sonja [21]
Answer: wavelength !!
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3 0
3 years ago
What is the difference between each distance traveled and displacement travled
-BARSIC- [3]

Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.

While, Distance is a scalar magnitude that measures over the trajectory.

8 0
3 years ago
car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0
3 years ago
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