1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Jet001 [13]
3 years ago
7

After a projectile is fired into the air, what is the acceleration in the x-direction? (Assume no air resistance.)

Physics
2 answers:
storchak [24]3 years ago
5 0

Answer:

a_x = 0

Explanation:

When projectile is in air then the force on the projectile is only due to its weight

It is vertically downwards

F_g = mg(-\hat j)

so here since we don't have any air drag or friction due to air so there will be no force along its tangent or along x direction

so the acceleration along x direction will be zero

a_x = 0

a_y = - g

Gre4nikov [31]3 years ago
4 0
Without air resistance, there is no horizontal force on the projectile,
so it has no horizontal acceleration.  The answer is "zero".
 
You might be interested in
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball
jonny [76]

(a) The impulse delivered to the ball by the racket is 5.24 kg.m/s

(b) The work that the racket does on the ball is -35.1 Joule

<h3>Further Explanation</h3>

<u>Given :</u>

mass of ball = m = 0.06 kg

initial velocity = v₁ = -50.4 m/s

final velocity = v₂ = 37.0 m/s

<u>Unknown :</u>

(a) Impulse = I = ?

(b) Work = W = ?

<u>Solution :</u>

<h2>Question (a) :</h2>

In this question , we could use the formula from Second Law of Newton :

I = \Delta p

I = p_2 - p_1

I = m \times v_2 - m \times v_1

I = m \times (v_2 - v_1)

I = 0.06 \times (37.0 - (-50.4))

I = 0.06 \times (87.4)

I = 5.244~kg.m/s

\large { \boxed {I \approx 5.24~kg.m/s} }

<h2>Question (b) :</h2>

W = F \times d

W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)

W = \frac{I(v_1 + v_2)}{2}

W = \frac{5.244(-50.4 + 37)}{2}

W = \frac{5.244(-13.4)}{2}

W = -35.1348~Joule

\large { \boxed {W \approx -35.1~Joule}}

<h3>Learn more</h3>

Newton's Law of Motion: brainly.com/question/10431582

Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Newton, Law, Impulse, Work

6 0
4 years ago
If the net force acting on a stationary object is zero, then the object will A. remain at rest. B. accelerate in the direction o
ra1l [238]

The object will remain at rest

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

Here net force is zero

An object with a net force of zero acting on it will remain at rest, if initially at rest, or it will maintain a constant velocity.

F=MA

Force F=0

Than Acceleration A=0

With zero Acceleration stationary object  will remain at rest.

Hence Remain at rest is the correct answer

Learn more about Newton's law of motion here

brainly.com/question/25545050

#SPJ4

6 0
2 years ago
This is the type of reaction that occurs when an acid cancels with a base.
viva [34]
Neutral reaction is the reaction that occurs when an acid cancels with a base
7 0
3 years ago
Read 2 more answers
Jessica's weight is 540 newtons. According to the Third Law, if the Earth pulls on Jessica with a force of 540 N, then Jessica p
Alja [10]

This is because the earth is way heavier than Jessica hence has a greater force of gravity applied on Jessica that Jessica applies to earth. The distance between the Earth and Jessica is also very short hence the force of gravity on Jessica towards earth is greater. According to Newton's laws that states; if an unbalanced force acts on an object, it will change the object's state of motion (the object will accelerate). The earth cause Jesica to ‘accelerate’ towards the ‘core’ when she stumbles.

8 0
3 years ago
Read 2 more answers
Consider the baby being weighed in Figure 4.25.
Lunna [17]

The mass and tension due to the system are as follows:

  • The mass of the child and scale = 10.6 kg
  • The tension T, in the cord attaching the child to the scale = 104N
  • The tension T', in the cord attaching the scale to the ceiling T' = 108.9 N

<h3>What is tension?</h3>

Tension is a type of pulling force due transmitted by means of a string or cable.

<h3>Force = mass * acceleration due to gravity</h3>

a) The mass of the child and scale = 104/9.81 = 10.6 kg

b) The tension T, in the cord attaching the child to the scale = scale reading = 104N

c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale

T' = 104 + (0.5 * 9.81)

T' = 108.9 N

d) The sketch is attached in the picture

In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.

Learn more about tension force at: brainly.com/question/24994188

#SPJ1

4 0
2 years ago
Other questions:
  • The law of reflection establishes a definite relationship between the angle of incidence of a light ray striking the boundary be
    13·1 answer
  • Consider a deer that runs from point A to point B. The distance the deer runs can be greater than the magnitude of its displacem
    10·1 answer
  • When playing a game of disc golf, each throw to your target is considered to be a O Point O Stroke O Hit O Toss​
    11·1 answer
  • How fast will a mass of water dropped from an airplane be moving after 3 seconds in the alr? (Ignore air resistance.)
    12·1 answer
  • When you measure the mass of an object you are also measuring its what
    10·1 answer
  • A real image can be obtained with:
    7·2 answers
  • Why are streams and rivers important for ecosystems?
    7·2 answers
  • I WILL MARK BRAINLIEST IF YOU ANSWER THIS PHYSICS QUESTION!!!
    12·1 answer
  • The source of the hot, intergalactic gas in many rich, regular clusters of galaxies appears to be.
    11·1 answer
  • What is electronic logic​
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!