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Anika [276]
2 years ago
7

If you know all of the forces acting on a moving object, can you tell in which direction the object is moving? if the answer is

yes, explain how. If the answer is no, give an example.
Physics
1 answer:
Montano1993 [528]2 years ago
8 0

No, knowing all the forces is not enough to know the direction of motion.

<h3>How to use Newton's laws?</h3>

The second Newton's law states that:

F = m*a

This says <u><em>"force equals mass times acceleration".</em></u>

Where acceleration is the rate of change of the speed. From that equation, we conclude that the acceleration is in the same direction that the net force.

So, if we know all the forces acting on an object, we know the net force acting on it, then we know the direction of the acceleration.

<h3>Is this enough to know the direction in which it is moving?</h3>

No, the object does not need to move in the same direction than its acceleration, the direction of motion will also depend on the initial velocity of the object (if it is initially moving with constant speed).

If we don't know that, we can't find the direction of motion.

An example of this can be a car going at 100km/h east-wise.

Then we apply a net force due west, then we have an acceleration due west. But as the initial direction of motion was east, the car will still move to the east, but the velocity will decrease gradually.

So as you can see in that example, we need to know the initial velocity to know the direction in which the object is moving.

If you want to know more about acceleration, you can read:

brainly.com/question/605631

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Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up
kirza4 [7]

Answer:

I believe the answer is b

Explanation:

5 0
3 years ago
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           N_x=86.62N

  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

                           N_y=F_V=mg-Tsin59\\

  • To find Ny, we need to find the tension T.
  • For this, we can equate the net horizontal force.

                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

  • Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

                    N_y= (40*9.8)-(169.8*sin59)=246.4N

  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

4 0
1 year ago
Read 2 more answers
active and passive secruity measures are employed to identify, detect, classify and analyze possible threats inside of which zon
Juli2301 [7.4K]

Answer:

Assessment zone

Explanation:

It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.

8 0
3 years ago
You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend
Bogdan [553]

Answer:

a.\tau=200J b.\alpha=0.44 \frac{rad}{s^2} c. \alpha=0.33\frac{rad}{s^2} d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by \tau=rF\sin \theta, being r the radius, F the force aplied and \theta the angle between the vector force and the vector radius. Since \theta=90^{\circ}, \, \sin\theta=1 and so \tau=rF=2m100N=200Nm=200J.

b. Since the relation \tau=I\alpha hols, being I the moment of inertia, the angular acceleration can be calculated by \alpha=\frac{\tau}{I}. Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is I=\frac{1}{2}Mr^2, being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by I_p=m_pr_p^{2}, being m_p the mass of the person and r_p^{2} the distance from the person to the center. Given all of this, we have

\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}.

c. Similar equation to b, but changing r_p=2m, so

alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}.

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0
3 years ago
When making a turn, do not have the steering wheel turned in the direction of the turn before beginning the turning maneuver.a)
RideAnS [48]

Answer:

a) True.

Explanation:

If you turn the wheel in the direction of the turn before beginning the turning maneuver then it's possible that there might be not enough space available for turning and also if you are waiting for the traffic to get clear with rear  ended then it will get pushed forward onto the coming traffic.

7 0
3 years ago
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