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4 years ago

When two objects collide, the momentum of the system remains constant. This is a statement of ______________.

Physics

2 answers:

Gennadij [26K]4 years ago

7 0

Answer:

B. Newton's third law

Explanation:

When two bodies collide the moment remains constant.

This occurs because we define a system that is formed by the two bodies, so that the forces last = before the shock are forces of action and reaction, from one body to the other, so that for the system (two bodies) the force total is zero, we assume that the system is isolated from the outside.

In summary, the initial establishment is a consequence of Newton's third law

Fudgin [204]4 years ago

3 0

Conservation of momentum

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Human populations all over the world burn fossil fuels to power cars, planes, and generate electricity. Select the statement tha

Eddi Din [679]

A.) Burning fossil fuels pollutes the environment and our heavy reliance on them creates a long-term problem because they are not a renewable resource.

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4 years ago

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What is the mass of a cannonball if a force of 2,500 N gives the cannonball an acceleration of 200m/s2?

pentagon [3]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

Divide each side
by 'acceleration': Mass = (force) / (acceleration)

   = (2,500 N) / (200 m/s²)

   = 12.5 kg

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4 years ago

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When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static

natali 33 [55]

The coefficient of static friction is 0.234.

Answer:

Explanation:

Frictional force is equal to the product of coefficient of friction and normal force acting on any object.

So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.

Normal force = mass * acceleration due to gravity

Normal force = 2 * 9.8 = 19.6 N.

And the frictional force is given as 4.6 N, then

$Coefficient of static friction = Frictional force/Normal force$

Coefficient of static friction = 4.6 N / 19.6 N = 0.234

So the coefficient of static friction is 0.234.

3 0

3 years ago

Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance

SSSSS [86.1K]

Answer:

${F_{tot} = -1.092*10^{-2}N$

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

Your answer may be positive or negative, depending on the direction of the force.

Solution:

The coulomb force is given by the equation

$F =k\dfrac{q_1 q_2}{d^2}.$

where $d$ is the separation between the charges $q_1$ and $q_2$ .

Now, in our case

$q_1=-13.5nC=-13.5*10^{-9}C$

$q_2=-1.735nC=-1.735*10^{-9}C$

$q_3= 47*10^{-9}C$

The separation between charges $q_3$ and $q_1$ is

$d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m$

Therefore, the force between them is

$F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N$

and it is directed in the negative x-direction.

The separation between charges $q_2$ and $q_3$ is

$d_{23} =-1.240mm-0=-1.240*10^{-3}m$

therefore, the force between them is

$F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N$

Therefore the total force on charge $q_3$ is

$F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}$

8 0

3 years ago

An object is traveling at 15 m/s and then after 5 seconds ends up traveling at 10 m/s. Calculate the speeds acceleration rate. W

SSSSS [86.1K]

Answer:

1) a = -1 m/s²

2) v = 12 m/s

Explanation:

Given,

The initial velocity of the object, u = 15 m/s

The final velocity of the object, v = 10 m/s

The time taken by the object to travel is, t = 5 s

Using the first equation of motion

v = u + at

a = (v - u) / t

Substituting the values

a = (10 - 15) / 5

= -1 m/s²

The negative sign indicates the body is decelerating

The acceleration of the object is, a = -1 m/s²

The speed of the object after 2 seconds

From the above equations of motion

v = 15 + (-1) 2

= 12 m/s

Hence, the speed of the object after 2 seconds is, v = 12 m/s

7 0

4 years ago