Answer: 11 m/s
vinitial=2 m/s
time=3 s
acceleration = 3 m/s^2
vfinal = ?
The key here is that it is a constant acceleration, so we can use the constant acceleration equations. The easiest one to use would be:
vfinal=vinitial + a*t
We need vfinal, so algebraically we are ready to put in numbers into the equation:
vfinal=vinitial + a*t = 2 m/s + (3 m/s^2)*(3 s ) = 11 m/s is the final velocity
Explanation:
Below is an attachment containing the solution.
Answer:
the answer is static electricity
Answer:
The location of the shear center o is 0.033 or 33 m
Explanation:
Solution
Recall that,
The moment of inertia of the section is = I = 0.05 * 0.4 ^3 /12 + 0.005 * 0.2 ^3/12
= 30 * 10 ^ ⁻⁶ m⁴
Now,
The first moment of inertia is
Q =ῩA = [ (0.1 -x) + x/2] (0.005 * x)
= 0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²
Thus,
The shear flow is,
q = VQ/I
so,
P = (0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²)/ 30 * 10 ^⁻⁶
P = (16.67 x - 83. 33 x²)
The shear force resisted by the shorter web becomes
Vw,₂ = 2∫ = ₀.₁ and ₀ = P (16.67 x - 83. 33 x²) dx = 0.11x
Then,
We take the moment at a point A
∑Mₐ = 0
- ( p * e)- (Vw₂ * 0.3 ) = 0
e = 0.11 p * 0.3/p
which gives us 0.033 m
= 33 m
Therefore the location of the shear center o is 0.033 or 33 m
Note: Kindly find an attached diagram to the question given above as part of the explanation solved with it.
To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

Where,
F = Force
r = Distance
Replacing,


The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

I = Moment of inertia
= Angular acceleration
Replacing with our values we have that



Therefore the value of moment of inertia is 