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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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A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will
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Answer:

The change  in momentum is  \Delta p =   kg \cdot m/s      

Explanation:

From the question we are told that  

       The mass of the probe is  m = 170 kg

       The location of the prob at time t = 22.9 s is  A  =

       The  momentum at time  t = 22.9 s is  p = < 51000, -7000, 0> kg m/s

        The net force on the probe is  F =  N

Generally the change in momentum is mathematically represented as

              \Delta p = F * \Delta t

The initial time is   22.6 s

 The final time  is  22.9 s

             Substituting values  

           \Delta p =  * (22.9 - 22.6)

            \Delta p =  * (0.3)  

              \Delta p =   kg \cdot m/s        

 

6 0
3 years ago
Describe three pieces of evidence that support the Big Bang theory
lisov135 [29]

Explanation :

According to astronomers, the whole universe is started with a giant explosion called as Big Bang. Big Bang theory shows that the universe is  extended from high density state.

There are some evidence for big bang as :

(1) There are some red shifts of different galaxies which means that the universe is expanding.  

(2) Due to the expanding of universe, some of the new elements are created like hydrogen, deuterium etc.

(3) Microwaves are detected by orbiting detectors.

All this parameters shows that big bang theory was correct.

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3 years ago
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2 years ago
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The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

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