Answer:
8 N North.
Explanation:
Given that,
One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.
We need to find the magnitude of net force acting on the object.
Let North is positive and South is negative.
Net force,
F = 10 N +(-2 N)
= 8 N
So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".
Answer:
The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V
Explanation:
Given:
(A)
Current 
A
Voltage 
V
For finding the resistance,
12Ω
(B)
For finding power delivered,
Watt
(C)
For finding the potential difference,
V
Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V
That can be considered alive.
T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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Of the forces listed I think the force of him diving and sliding across the infield acted on the player.
I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.
