If the box is a distance 1.81 m from the rear of the truck when the truck starts,<span> ... Force of Friction = mu_s * Normal Force( </span>M<span> * G) ... The </span>box starts<span> moving! ... Now that the </span>box<span> is moving, the bed of the </span>truck<span> pulls at it with 17.4 ... out how </span>long<span> it will take the </span>box<span> to reach the back of the </span>truck<span>. ... T^2 = 2 * </span>1.81<span> / .64</span>
Answer:
Student A is correct
Explanation:
Colored objects look the way they do because of reflected light. When sunlight is shined on a green leaf, the violet, red and orange wavelengths are absorbed. The reflected wavelengths appear green. In each case we are seeing the complementary colors to the ones absorbed.
Answer:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
Explanation:
For this exercise we use the definition moment of inertia
I = ∫ r² dm
for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are
at one end
I₁ = ⅓ mL²
in in center
I₂ = 
m L²
There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder
I₃ = 
m r²
remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small
when examining the different moments of inertia:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
Answer:
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ
