Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Answer:
f(x)=a(x - h)2 + k
Much like a linear function, k works like b in the slope-intercept formula. Like where add or subtract b would determine where the line crosses, in the linear, k determines the vertex of the parabola. If you're going to go up 2, then you need to add 2.
The h determines the movement horizontally. what you put in h determines if it moves left or right. To adjust this, you need to find the number to make the parentheses equal 0 when x equals -2 (because moving the vertex point to the left means subtraction/negatives):
x - h = 0
-2 - h = 0
-h = 2
h = -2
So the function ends up looking like:
f(x)=a(x - (-2))2 + 2
Subtracting a negative cancels the signs out to make a positive:
f(x)=a(x + 2)2 + 2Explanation:
This question is based on the fundamental assumption of vector direction.
A vector is a physical quantity which has magnitude as well direction for its complete specification.
The magnitude of a physical quantity is simply a numerical number .Hence it can not be negative.
A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector. For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.
As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.
The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.
Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.
In a general sense we assume that vertically downward motion is negative and vertically upward is positive. In case of a falling object the motion is vertically downward. So the velocity of that object is negative .
So last option is partially correct as the vector can be negative depending on our choice of co-ordinate system.
