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Vilka [71]
4 years ago
8

A speedboat moves on a lake with initial velocity vector v1,x = 8.57 m/s and v1,y = -2.61 m/s, then accelerates for 6.67 seconds

at an average acceleration of aav,x = -0.105 m/s2 and aav,y = 0.101 m/s2. What are the components of the speedboat\'s final velocity, v2,x and v2,y?
Find the speedboat's final speed?
Physics
1 answer:
KengaRu [80]4 years ago
3 0
First, you find the velocity at each component. The general equation is:

a = (v2 - v1)/t

a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s

a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s

To find the final speed, find the resultant velocity by taking the hypotenuse.

v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s
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A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and
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The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\

Where:

  • mb is the mass of the bowling ball
  • mp the mass of the pin
  • v_{0b}\quad and\quad v_{0p} the initial velocities of the bowling ball and the pin.
  • v_{1b}\quad and\quad v_{1p} the final velocities of the bowling ball and the pin.

Solving for v0b:

v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}

<h2>R/ The original velocity of the ball was 5.71 m/s.</h2>
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3 years ago
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Answer:

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Explanation:

Given that,

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We need to find the acceleration of the motorbike at this moment in time.

Net force acting on the bike is given by :

F = 2000 N - 845 N = 1155 N

Let a be the acceleration of the motorbike. So,

a=\dfrac{F}{m}\\\\a=\dfrac{1155}{286}\\\\a=4.03\ m/s^2

So, the acceleration of the motorbike is 4.03\ m/s^2.

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Answer:

B

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B

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