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3 years ago

Number 10 and an explaination would be fabulous. thanks!

Physics

2 answers:

Sati [7]3 years ago

3 0

Linear momentum has to be conserved. It was zero before the thread eas burned ... when nothing was moving ... so the momentum of the masses moving in opposite directions has to add up to zero. ... Momentum = mass times speed. ... In one direction, you have 5 kg times 1/5 m/s= 1 kg-m/s. ... We need 1 kg-m/s in the other direction. ... 7 kg times speed = 1 kg-m/s. ... Can you finish it from here ?

UkoKoshka [18]3 years ago

3 0

The answer would be 1/7 m/s because the initial momentum in the objects is 0, then the final momentum of the objects must be 0. if the 5 kg object has a momentum of 1 (5 * 1/5) then the other object must be -1. The only velocity that will produce that result is -1/7 m/s, but since speed doesnt take into account the direction of an object.

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What is the charge on an object that has 1.09x10^13 excess electrons?

Nadya [2.5K]

Answer:

$-1.74\cdot 10^{-6} C$

Explanation:

The electron is the particle that rotates around the nucleus of the atom; it has a negative electric charge equal to :

$e=-1.6\cdot 10^{-19}C$

which is known as fundamental charge.

For an object containing N excess electrons, the total charge of the object is

$Q=Ne$

In this problem, the number of excess electrons in the object is:

$N=1.09\cdot 10^{13}$

Therefore, by plugging it the numbers, we can find the value of Q, the total charge of the object:

$Q=(1.09\cdot 10^{13})(-1.6\cdot 10^{-19})=-1.74\cdot 10^{-6} C$

5 0

4 years ago

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