Answer:
Distance is 50m
Displacement is 0m
Explanation:
Distance is based on the amount of length you covered, regardless of where you end.
Displacement only considered where you started and where you ended, which is at the same spot in this case. Therefore, no displacement.
Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
- The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>
Reasons:
The energy given to the block by the spring =
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block = = kinetic energy of
block =
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring =
- The force of the weight of the block on the string,
The energy given to the block = = The kinetic energy of block as it leaves the spring =
Which gives;
Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and <em>c</em> are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
- As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.
<em>Please find attached a drawing related to the question obtained from a similar question online</em>
<em>The possible question options are;</em>
- <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
- <em>The relationship is no longer linear and v will be more for the same value of x</em>
- <em>The relationship is still linear, with lesser value of v</em>
- <em>The relationship is still linear, with higher value of v</em>
- <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>
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Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
A quadrilateral with only one pair of parallel sides.
<span>a small carpal bone in the base of the hand, articulating with the metacarpal of the index finger.
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