Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
The answer is as voltage increases current increases and therefore resistance would remain constant
To find:
The equation to find the period of oscillation.
Explanation:
The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.
Thus the period of a pendulum is given by the equation,
Where L is the length of the pendulum and g is the acceleration due to gravity.
On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.
Final answer:
The period of oscillation of a pendulum can be calculated using the equation,
Answer:
5x10^-3
Explanation:
Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.
Hooke's Law can be represented as
<h3> F = kx, </h3>
<em>where F is the force </em>
<em> k is the spring constant</em>
<em> x is the extension of the material </em>
<em />
Plug values in the equation
Step 1 find the original extension
0.045 = (400)x
x = 1.125x 10^-4 m d
Step 2 find the new extension
0.045+2 = 400(x)
2.045 = 400x
x = 5.1125x10^-3
Step 3 subtract the new extension with original
Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3
