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

Answer:
Net electric field, 
Explanation:
Given that,
Charge 1, 
Charge 2, 
distance, d = 3.2 cm = 0.032 m
Electric field due to charge 1 is given by :



Electric field due to charge 2 is given by :



The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :



So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.
Answer:
(c) +6.67
Explanation:
f1 = 10 cm
f2 = 20 cm
u = Object distance = 15 cm
Distance between lenses = 20 cm
For first lens image distance

Distance from second lens is 10 cm to the right

The final image will appear as +6.67 cm
Answer: 9130 joules
Explanation:
Workdone by wheelbarrow = ?
Time = 11 seconds
Power = 830 watts
Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.
i.e Power = (workdone/time)
830 watts = Workdone / 11 seconds
Workdone = 830 watts x 11 seconds
Workdone = 9130 joules
Thus, 9130 joules of work is required to get the wheelbarrow across the yard.
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s
Let h₁ = the height at time t, for t ≤ 3.98 s
Let h₂ = the height at time t > 3.98 s
Motion for t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s
Motion for t > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m
The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m
Answer: 931.4 m (nearest tenth)