Answer:
"Magnitude of a vector can be zero only if all components of a vector are zero."
Explanation:
"The magnitude of a vector can be smaller than length of one of its components."
Wrong, the magnitude of a vector is at least equal to the length of a component. This is because of the Pythagoras theorem. It can never be smaller.
"Magnitude of a vector is positive if it is directed in +x and negative if is is directed in -X direction."
False. Magnitude of a vector is always positive.
"Magnitude of a vector can be zero if only one of components is zero."
Wrong. For the magnitude of a vector to be zero, all components must be zero.
"If vector A has bigger component along x direction than vector B, it immediately means, the vector A has bigger magnitude than vector B."
Wrong. The magnitude of a vector depends on all components, not only the X component.
"Magnitude of a vector can be zero only if all components of a vector are zero."
True.
Answer:
c. because A will land first becuase its heavier :)
Explanation:
Answer:
yes
Explanation:
you will feel weary after shorter times
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 =
/ W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s