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AleksandrR [38]
2 years ago
13

In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into

castles. You are asked to study the motion of such a projectile for a group of local enthusiast planning a medieval war reenactment. Unfortunately an actual trebuchet had not been built yet, so you decide to first look at the motion of a thrown ball as a model of rocks thrown by a trebuchet. Specifically, you are interested in how the horizontal and the vertical components of the velocity for a thrown object change with time. 1. Make a large rough sketch of the trajectory of the ball after it has been thrown. Draw the ball in at least five different positions; two when the ball is going up, two when it is going down, and one at its maximum height. Label the horizontal and vertical axes of your coordinate system.
2. On the sketch, draw and label the expected acceleration vectors of the ball (relative sizes and directions) for the five different positions. Decompose each acceleration vector into its vertical and horizontal components.
3. On the sketch, draw and label the velocity vectors of the object at the same positions you chose to draw your acceleration vectors. Decomposes each velocity vector into its vertical and horizontal components. Check to see that the changes in the velocity vector are consistent with the acceleration vectors.
4. Looking at the sketch, how does someone expect the ball's horizontal acceleration to change with time? Could you give a possible equation giving the ball's horizontal acceleration as a function of time? Graph this equation. If there are constants in your equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
5. Looking at the sketch, how does someone expect the ball's horizontal velocity to change with time? Is it consistent with the statements about the ball's acceleration from the previous question? Could you give a possible equation for the ball's horizontal velocity as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
6. Could you give a possible equation for the ball's horizontal position as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph? Are any of these constants related to the equations for horizontal velocity or acceleration?
7. Repeat questions 4-6 for the vertical component of the acceleration, velocity, and position. How are the constants for the acceleration, velocity and position equations related?
Physics
1 answer:
NNADVOKAT [17]2 years ago
4 0

Answer:

2) a_y= -g  3) vₓ=constant v_y = v_{oy} - g t, 4)  vₓ = v₀ₓ - ax t

5)  changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

           v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

           vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

           x = v₀ₓ t

           y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

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(a) what is the system of interest if the acceleration of the child in the wagon is to be calculated? (select all that apply.)
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since child is moving along with the wagon and we need to find the acceleration of child inside that wagon then in this case the system of interest must be child + wagon

System of interest will be the system that is used to find the force or acceleration using Newton's law

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6 0
2 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

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v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
A portion of the mesosphere and thermosphere known for its ability to "bounce" radio signals is the _____________
Lerok [7]

It is B: ionosphere. This has the ability to bounce radio signals
8 0
2 years ago
Identify the person who made the correct statement.
dmitriy555 [2]
The answer is:  [C]:  Neither Juan nor Christina are correct.
_________________________________________________
Explanation:
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This statement is incorrect.  There are many instances of fossil deposits containing more than one type of organism.
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Consider Christina's statement.
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"Christina said fossil deposits never contain one type of organism."
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This statement is incorrect.  Fossil deposits, by definition (of "fossil") ALWAYS contain <u>at least one type</u> of organism.
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As such, neither Juan nor Christina are correct. 
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Answer choices A, B, and D are ruled out.
7 0
2 years ago
A pitcher claims he can throw a 0.148-kg baseball with as much momentum as a 2.00-g bullet moving with a speed of 1.50 ✕ 103 m/s
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(a) The mass of the bullet is m_b = 2.00g=0.002 kg and its speed is v_b = 1.5 \cdot 10^3 m/s, so its momentum is
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The pitcher claims that he can throw the ball with the same momentum p of the ball. Since the mass of the ball is m=0.148 kg, this means that the velocity of the ball must be:
v= \frac{p}{m}= \frac{3 kg m/s}{0.148 kg}=20.3 m/s

(b) The kinetic energy of the bullet is:
K_b =  \frac{1}{2} m_b v_b^2= \frac{1}{2}(0.002 kg)(1.5 \cdot 10^3 m/s)^2=2250 J

while the kinetic energy of the ball is:
K= \frac{1}{2}mv^2= \frac{1}{2}(0.148 kg)(20.3 m/s)^2=30.5 J

So, the bullet has greater kinetic energy than the ball.
4 0
2 years ago
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