Answer:
12345
Explanation:
yan na po answer ko hehehe
Answer:
As collision is elastic,thus we can use conservation of momentum equation
mA=0.2 kg
(vB)1=0 m/s.......................as it is on rest before collision
(vA)1=4 m/s
(vA)2=-1 m/s
(vB)2=2 m/s
using equation
(mA*vA+mB*vB)1= (mA*vA+mB*vB)2
Where 1 and 2 represents before and after collision
(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)
0.8=-0.2+(2mB)
mass of object B=mB=0.3 Kg
If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:
<h3>How is the ratio of the perpendicular parts obtained?</h3>
To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle = 4m.
The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:
r = √m/e : √2m/e : √4m/2e
When resolved, the resulting ratios will be:
1: √2 : 1
Learn more about the radii of their circular paths here:
brainly.com/question/16816166
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You would be correct.
Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.
Hope this helps!