<span>A differences in the warmth, or moisture level as well as neighbouring areas of pressure in </span>air cause air<span> to circulate. in the earth's atmosphere.</span>
Answer:
A. the magnitude of the velocity at which the two players move together immediately after the collision is 7.9m/s
B. The direction of this velocity is due north as the linebacker since he has obviously has more momentum
Explanation:
This problem bothers on the inelastic collision
Given data
Mass of linebacker m1= 110kg
Mass of halfbacker m2= 85kg
Velocity of linebacker v1= 8.8m/s
Velocity of halfbacker v2= 7.2m/s
Applying the principle of conservation of momentum for inelastic collision we have
m1v1 +m2v2= (m1+m2)v
Where v is the common velocity after impact
Substituting our data into the expression we have
110*8.5+85*7.2= (110+85)v
935+612=195v
1547=195v
v=1547/195
v=7.9m/s
Momentum of linebacker after impact = 110*7.9= 869Ns
Momentum of halfbacker after impact = 85*7.9= 671.5Ns
the direction after impact is due north since the linebacker has greater momentum
Answer:
.5 m N
Explanation:
Displacement just means distance from starting point, so moving 3 m north then 2.5 m south is the same as if you took 3 steps forward then 2.5 steps backward. The end point is only .5 away from the start point.
Answer:
Explanation:
Generally the workdone in moving the proton is mathematically represented as
Where 
So
Here 
is the velocity at A with value 50 m/s
So
Also
Here is the velocity at A with value 
=> 
=> 
So
Now this workdone is also mathematically represented as
So
Here 
So
Generally proton movement is in the direction of the electric field it means that 
So
Direct rotation is your answer .-.
