Answer:
(a) The total energy of the object at any point in its motion is 0.0416 J
(b) The amplitude of the motion is 0.0167 m
(c) The maximum speed attained by the object during its motion is 0.577 m/s
Explanation:
Given;
mass of the toy, m = 0.25 kg
force constant of the spring, k = 300 N/m
displacement of the toy, x = 0.012 m
speed of the toy, v = 0.4 m/s
(a) The total energy of the object at any point in its motion
E = ¹/₂mv² + ¹/₂kx²
E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²
E = 0.0416 J
(b) the amplitude of the motion
E = ¹/₂KA²

(c) the maximum speed attained by the object during its motion

<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
Answer:
3.31m/s
Explanation:
Angular momentum for 3s is



Moment if inertia is


Angular speed
ω = L/I

The speed of each ball is
V = ωL

Explanation:
Formula which holds true for a leans with radii
and
and index refraction n is given as follows.
Since, the lens is immersed in liquid with index of refraction
. Therefore, focal length obeys the following.
and,
or,
= 32.4 cm
Using thin lens equation, we will find the focal length as follows.

Hence, image distance can be calculated as follows.


= 47.9 cm
Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.
Answer:
2.25in³
Explanation:
For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³
See attached Table 314.16(B) from NEC 2011