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3 years ago

Provide an example of when momentum is conserved and explain your answer you can get 10 PTS if answered with a good explaination

Physics

1 answer:

dezoksy [38]3 years ago

4 0

Answer:

$m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s$

Explanation:

<u>Conservation of Momentum </u>

The total momentum of a system of two particles is

$p=m_1v_1+m_2v_2$

Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now

$p'=m_1v_1'+m_2v_2'$

The momentum is conserved if no external forces are acting on the system, thus

$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

Let's put some numbers in the problem and say

$m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s$

$(8)(12)+(6)(4)=(8)(-6)+(6)(28)$

$96+24=-48+168$

120=120

It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s

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A negatively charged object has?

mixas84 [53]

More, fewer electrons than object is neutral

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3 years ago

Please someone derive 3rd equation of motion u- v= 2sa

zlopas [31]

We know

$\\ \sf\longmapsto a=\dfrac{dv}{dt}$

$\\ \sf\longmapsto a=\dfrac{dv}{dx}.\dfrac{dx}{dt}$

$\\ \sf\longmapsto a=v\dfrac{dv}{dx}$

$\\ \sf\longmapsto adx=vdv$

Integrate

$\\ \sf\longmapsto a{\displaystyle{\int}^x_{x_0}}dx=\displaystyle{\int}_u^v vdv$

$\\ \sf\longmapsto a(x-x_0)=\dfrac{v^2-u^2}{2}$

Here

x-x_0=s

$\\ \sf\longmapsto v^2-u^2=2as$

6 0

3 years ago

Read 2 more answers

You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?

Inga [223]

Answer:

The max height of the ball is y = -1/2 (v0²/g).

It takes the ball t = -2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = -v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t

When the ball is at max height, the velocity is 0. So, let´s find the time at which the velocity of the ball is 0.

v = v0 + g · t

0 = v0 + g · t

t = -v0/g

Now, replacing t = -v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g · (-v0/g)²

y = -(v0²/g) + 1/2 · (v0²/g)

y = -1/2 (v0²/g)

The max height of the ball is y = -1/2 (v0²/g). Remember that g is negative.

Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = -2 v0/g.

However, let´s calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · t + 1/2 · g · t²

0 = t · ( v0 + 1/2 · g · t)

0 = v0 + 1/2 · g · t

-2 · v0/g = t

It takes the ball t = -2 · v0/g to hit the ground.

Let´s use the equation of velocity at final time (t = -2 · v0/g):

v = v0 + g · t

v = v0 + g · ( -2 · v0/g)

v = v0 - 2· v0

v = -v0

The speed of the ball when it hits the ground is v = -v0.

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4 years ago

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uranmaximum [27]

Answer:

1.7531 smu/yn

Explanation:

check the attached file below for answer and explanation.

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3 years ago

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rewona [7]

It's shows static electricity because of the hair

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