____ charges are attract each other

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

2 years ago

Newton's third law can be summarized as "every action has an equal and opposite reaction". In this problem, consider the action

scoundrel [369]

Answer:

Action force: Would be the force of your feet against the Earth given by the weight defined as:

$W = mg$

Where g is a constat who represent the gravity $g =9.8 m/s^2$ in Earth

Reaction force: Would be the force of the Earth pushing against your feet. And on this case is represented by the normal force defined as:

$N = \mu f_f$

Where $\mu$ represent the friction coefficient between the ground and the object.

And $f_f$ the friction force.

If we don't have any other forces involved in the y axis we can conclude that:

$W=N= mg$

And as we can see we have that Action force = Reaction force

So then the third Law of Newton is satisfied.

Explanation:

For this case we have this:

Action force: Would be the force of your feet against the Earth given by the weight defined as:

$W = mg$

Where g is a constat who represent the gravity $g =9.8 m/s^2$ in Earth

Reaction force: Would be the force of the Earth pushing against your feet. And on this case is represented by the normal force defined as:

$N = \mu f_f$

Where $\mu$ represent the friction coefficient between the ground and the object.

And $f_f$ the friction force.

If we don't have any other forces involved in the y axis we can conclude that:

$W=N= mg$

And as we can see we have that Action force = Reaction force

So then the third Law of Newton is satisfied.

7 0

3 years ago

_________ is the angle of incidence is equal to the angle of reflection

ExtremeBDS [4]

Law of reflection is the angle of incidence = angle of reflection.

8 0

2 years ago

While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the

olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0

2 years ago

Factors that affect acceleration due to gravity.<br>

hram777 [196]

Answer

Factors Affecting Acceleration Due to Gravity

(1) Position on the planetary surface, it is higher at the axis (poles),

(2) Mass of the planet, higher with planets of larger masses

(3) Distance between object and the centre of planet, the larger the distance, the smaller the acceleration due to gravity.

Mark me as Brainlist if this answer is helpful for you

8 0

2 years ago

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