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BartSMP [9]
3 years ago
10

The path a projectile takes is known as the Question 1 options: vertical component trajectory horizontal component parabola Ques

tion 2 (1 point) Which of the following is not an example of a projectile? Question 2 options: A baseball that has been pitched A rock being thrown over a bridge A person sitting in a chair, not moving A car driving off a cliff Question 3 (1 point) A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds. How high above the ground was the airplane when the supplies were dropped? (Hint: d = 1/2gt2; g = -9.8 m/s2) Question 3 options: 120 meters 490 meters 980 meters 49 meters Question 4 (1 point) As the supplies fell, they traveled 650 meters horizontally. How fast were the supplies and the airplane moving forward? Remember that it took 10 seconds for them to reach the ground. (Hint: v = d/t) Question 4 options: 65 m/s 6500 m/s 0.154 m/s 660 m/s Question 5 (1 point) A projectile has a vertical component and a horizontal component. Question 5 options: True False Question 6 (1 point) The shape of a projectile's trajectory is an ellipse. Question 6 options: True False Question 7 (1 point) Which of the following is not true about a vertical component? Question 7 options: The vertical component of a projectile changes due to gravity. The vertical component acts the same as an object in free fall. The vertical component always equals the horizontal component. Question 8 (1 point) During the course of trajectory, the horizontal component of the velocity is Question 8 options: constant increasing decreasing negative Question 9 (1 point) In the simulation above, as the projectile travels upward, how does the vertical velocity change? Question 9 options: Vertical velocity decreases. Vertical velocity increases. Vertical velocity remains the same. Vertical velocity is zero. Question 10 (1 point) In the simulation above, as the projectile travels downward, how does the vertical velocity change? Question 10 options: Vertical velocity decreases. Vertical velocity is zero. Vertical velocity remains the same. Vertical velocity increases.
Physics
2 answers:
Aneli [31]3 years ago
7 0
<h2>Answer to Q 1:</h2>

<u>The path a projectile takes is called </u><u>Trajectory</u>

<h2>Explanation:</h2>

According to Physics, in terms of projectile motion, trajectory is the curved path that an object follows when it travels in open space. It has two components x and y. X component shows the range while y component shows the height of the projectile.

<h2>Answer to Question 2  </h2>

<u>The odd choice out of these options is </u><u>A person sitting in a chair, not moving</u>

<h2>Explanation:</h2>

As we know that it is necessary for a projectile motion that some movement must be there. Therefore some movement is necessary in a curved path. Since the person sitting in a chair is not moving and he is in rest so he cannot be considered as moving because he is not changing his position with respect to an observer.

<h2>Answer to Question 3</h2>

<u>The distance will be </u><u>490 meters</u>

<h2>Explanation:</h2>

According to second equation of motion

S= vit + ½ gt²

Putting the values

S = 0 + ½ (9.8) (10)

S = 490 meters

<h2>Answer to Question 4 </h2>

<u>The speed will be </u><u>65m/s</u>

<h2>Explanation:</h2>

As we know that

Speed = Distance / Time

So putting the values

Speed = 650 / 10

Speed = 65 m/s  

<h2>Answer to Q 5:</h2>

<u>The given statement is</u><u> true</u>

<h2>Explanation:</h2>

A projectile motion is composed of two components. X component is responsible for showing the range of the projectile and it starts from the origin or starting point while on the other hand Y component is responsible for the height of the projectile. We plot the maximum height that a projectile can gain on the Y axis of a graph.

<h2>Answer to Q 6</h2>

<u>The given statement is</u><u> false</u>

<h2>Explanation:</h2>

There is a difference in the shape of parabola and ellipse. The shape of projectile graph is a parabola, not an ellipse. The equation of a projectile gives us two values at two different points.  Since it is a quadratic equation so that is why this type of motion gives parabola and not an ellipse.

<h2>Answer 7:</h2>

<u>The false statement is </u><u>vertical component is always equals to horizontal component.</u>

<h2>Explanation:</h2>

The horizontal component of a projectile is X and it is constant which means no force is acting on it whereas y component has a force of gravity as the body goes up or comes down so both of these components can never be the same.

<h2>Answer to Q 8:  </h2>

<u>During the course of trajectory,</u><u> the horizontal component of the velocity is constant.</u>

<h2>Explanation:</h2>

The horizontal component of trajectory is the X component. This component does not have any force upon it except the force of friction of earth and object. The force of friction is also acting in x axis that is why we can say that during the course of trajectory, the horizontal component of the velocity is constant.

<h2>Answer to Q 9  </h2>

<u>In the simulation above, </u><u>as the projectile travels upward, its vertical velocity decreases</u>

<h2>Explanation:</h2>

We know that for a body moving up or down is affected by gravity so whenever an object moves against the force of gravity, it’s acceleration and velocity decreases because the gravitational pull tends the object to stay it on earth surface.

<h2>Answer to Q 10</h2>

<u>In the simulation above, as the projectile travels downward, </u><u> the vertical velocity increases</u>

<h2>Explanation:</h2>

When a body moves up or down it is affected by gravitational pull. As the body comes towards the earth the gravitational acceleration increases with the rate of 9.8m/s². This acceleration is taken as positive and when the projectile hits the ground it is in its maximum speed.

Firdavs [7]3 years ago
3 0

1. Trajectory

The path a projectile is called a trajectory in physics. It has a vertical component and even makes a parabola, but if we are talking about physics, it is trajectory.

2. A person sitting in a chair

Projectiles can be defined as an object that is in flight. So it has to be in the air. Since a person sitting in a chair is not in flight, then it is NOT a projectile. (Unless you throw the person in the air while he is in the chair)

3. 490 meters

We have the formula and our given:

d = 1/2gt²

Just plug in the values to get your answer:

d = 1/2(-9.8m/s²)(10s)²

d = (-4.9m/s²)(100s²)

d = -490m

So since height is a scalar value, just take out the negative sign.

4. 65 m/s

Again we have our formula and given:

v=\dfrac{d}{t}

So we just plug in our values:

v=\dfrac{650m}{10s}

v=650m/s

5. True

A projectile, if you will notice its trajectory moves both horizontally and vertically. The horizontal motion is what we call the x-component and the vertical is called the y-component. This is what gives it its' curved path.

6. False

An ellipse is an oval-shaped path. A projectile does not move in a circular/oval path. It travels a curved path. It can be parabolic. It is curved but it does not follow a circular path.

7. The vertical component always equals the horizontal component

This is false. Vertical component is different from the horizontal component. The horizontal component is not influenced by gravity only the vertical component. So in short, the horizontal component is the same throughout, but the vertical component changes over time.

8. Constant

Like mentioned above, the horizontal movement is constant or it does not change. This is because a projectile is defined also as an object that is influenced solely by gravity.

9. Vertical velocity decreases

This is because the movement is against the pull of gravity. It will continue to decrease until it will eventually come to a stop and start to descend. As it descends it increases.

10. Vertical velocity increases

I guess explained it above. As the object descends, the vertical movement increases. This is why you can actually die at certain heights. It increases as the time in flight increases. So the longer in flight, the faster it will get.



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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2
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Answer:

Approximately 0.608 (assuming that g = 9.81\; \rm N\cdot kg^{-1}.)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

  • Let m represent the mass of this car.
  • Let r represent the radius of the circular track.

This answer will approach this question in two steps:

  • Step one: determine the centripetal force when the car is about to skid.
  • Step two: calculate the coefficient of static friction.

For simplicity, let a_{T} represent the tangential acceleration (1.90\; \rm m \cdot s^{-2}) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

The angular acceleration of this car can be found as \displaystyle \alpha = \frac{a_{T}}{r}. (a_T is the tangential acceleration of the car, and r is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity (u and v) to (tangential) acceleration a_{T} and displacement x:

v^2 - u^2 = 2\, a_{T}\cdot x.

The idea is to solve for the final angular velocity using the angular analogy of that equation:

\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta.

In this equation, \theta represents angular displacement. For this motion in particular:

  • \omega(\text{initial}) = 0 since the car was initially not moving.
  • \theta = \displaystyle \frac{\pi}{2} since the car travelled one-quarter of the circle.

Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

Let m represent the mass of this car. The centripetal force at this moment would be:

\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, m\, g.

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let \mu_s denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g.

The size of this force should be equal to that of the centripetal force when the car is about to skid:

\mu_s\, m\, g = \pi\, m\, a_{T}.

Solve this equation for \mu_s:

\mu_s = \displaystyle \frac{\pi\, a_T}{g}.

Indeed, the expression for \mu_s does not include any unknown letter. Let g = 9.81\; \rm N\cdot kg^{-1}. Evaluate this expression for a_T = 1.90\;\rm m \cdot s^{-2}:

\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

(Three significant figures.)

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a) 0.0028 rad/s

b) 23.68 m/s^2

c) 0 m/s^2

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

\omega = \frac{\theta}{t}

where

\theta is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

\omega = \frac{v}{r} (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

v=29,960 km/h is the linear speed, converted into m/s,

v=8322 m/s

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the  centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

a_r=\omega^2 r

where

\omega is the angular speed

r is the radius of the orbit

For the spaceship in the problem, we have

\omega=0.0028 rad/s is the angular speed

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

a_t=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in the linear speed

\Delta  t is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

v=8322 m/s

Therefore, its linear speed is not changing, so the change in linear speed is zero:

\Delta v=0

And therefore, the tangential acceleration is zero as well:

a_t=\frac{0}{\Delta t}=0 m/s^2

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