We can calculate the length of each spring by using the relationship:
where
F is the force applied to the spring
k is the spring constant
x is the length of the spring (measured with respect to its rest position)
Re-arranging the equation, we have
The force applied to both spring is F=60 N. Spring A has spring constant of k=4 N/m, therefore its length with respect to its rest position is
Spring B has spring constant of k=5 N/m, so its length with respect to its rest position is
Therefore, the correct answer is
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D.Spring A is 3 m longer than spring B because 15 – 12 = 3.</span>
Answer:
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.
Answer:
Fx = -7042.86N (Taking the right as positive sense of motion)
Fy = 2154.29N.
Explanation:
The full solution can be found in the attachment below. The approach to the solution involves the summation of the respective components of the momentum along the horizontal and vertical axis.
To the right is taken as positive sense of motion and the left as negative sense of motion.
The basic statement of Newtown's second law have been used which is
M(v2 – v1) = Ft
See the attachment below for the complete calculation procedure.
Answer:
The power dissipated in either one of the parallel resistors is 2 V
Explanation:
Given;
two parallel resistors, R₁ and R₂ = 2 ohms
The total resistance of the Two resistors of 2 ohms connected in parallel is;
when connected to another resistor of 1 ohm in series, the total resistance becomes;
Rt = R₁ + R₂
Rt = 1 + 1 = 2 ohms
Current in the circuit, I = voltage / total resistance
= 2 /2 = 1 A
the overall circuit has been resolved to series connection, and current flow in series circuit is constant.
Power = I²R
Thus, power dissipated in either one of the parallel 2 ohms resistors is;
Power = I²R = (1)² x 2 = 2 V
I don’t really know the answer cause I need more information about the question