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Rudiy27
3 years ago
14

A firecracker in a coconut blows the coconut into three pieces. two pieces of equal mass fly off south and west, perpendicular t

o each other, at 24 m/s . the third piece has twice the mass as the other two. you may want to review (page 270) . part a what is the speed of the third piece?

Physics
2 answers:
netineya [11]3 years ago
8 0
Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
m_{1}  v_{1i} + m_{2}  v_{2i} + m_{3}  v_{3i} = m_{1}  v_{1f} + m_{2}  v_{2f} + m_{3}  v_{3f}
Before the firecracker blows a coconut does not move, so left side is equal to 0:
0 = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}

We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
v_{3f} = -  \frac{m_{1}  v_{1f}  + m_{2} v_{2f} }{ m_{3} }

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s
This speed we can insert into formula for v3f:
v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\  \\ v_{3f} = - \frac{2m*33.9 }{ 2m }  \\  \\ v_{3f} = - 33.9m/s

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.
expeople1 [14]3 years ago
4 0

The speed of the third piece is about 17 m/s

\texttt{ }

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

mass of object 1 = m₁ = m

mass of object 2 = m₂ = m

velocity of object 1 = v₁ = -24 j

velocity of object 2 = v₂ = -24 i

mass of object 3 = m₃ = 2m

<u>Asked:</u>

velocity of object 3 = v₃ = ?

<u>Solution:</u>

We will use conservation of momentum formula to solve this problem:

m_1u_1 + m_2u_2 + m_3u_3 = m_1v_1 + m_2v_2 + m_3v_3

0 + 0 + 0 = m(-24 \widehat{j}) + m (-24 \widehat{i}) + 2mv_3

0 = -24 \widehat{j} -24 \widehat{i} + 2v_3

2v_3 = 24 \widehat{i} + 24 \widehat{j}

v_3 = (24 \widehat{i} + 24 \widehat{j}) \div 2

\large {\boxed {v_3 = 12 \widehat{i} + 12 \widehat{j}} }

\texttt{ }

<em>We could calculate the magnitude of the vector of velocity of this third piece as follows:</em>

|v_3| = \sqrt{12^2 + 12^2}

|v_3| = \sqrt{2 (12^2)}

|v_3| = 12\sqrt{2} \texttt{ m/s}

|v_3| \approx 17 \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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S_A_V [24]

Answer:

1.59 seconds

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but if you are wise you will read the entire answer.

Explanation:

This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.

Time

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t = 20 / 12.6

t = 1.59 seconds.

It takes 1.59 seconds to hit the ground

Height of the building

<u>Givens</u>

t = 1.59 sec

vi = 0     Another critical point. The beginning speed vertically is 0

a = 9.8 m/s^2   The acceleration is vertical.

<u>Formula</u>

d = vi*t + 1/2 a t^2

<u>Solution</u>

d = 1/2 a*t^2

d = 1/2 * 9.8 * 1.59^2

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7 0
3 years ago
A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to
gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

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So, indeed, after 10 seconds, the velocity reaches 1000 m/s

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Answer:

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Explanation:

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Δy = -175 m

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t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

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Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

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