A 2.0\, \text {kg}2.0kg2, point, 0, start text, k, g, end text cart moving right at 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction on a frictionless track collides with a 3.0\,\text {kg}3.0kg3, point, 0, start text, k, g, end text cart initially at rest. The 2.0\, \text {kg}2.0kg2, point, 0, start text, k, g, end text cart has a final speed of 1.0\,\dfrac{\text m}{\text s}1.0 s m ​ 1, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the left. What is the final speed of the 3.0\,\text {kg}3.0kg3, point, 0, start text, k, g, end text cart?

.20

Step-by-step explanation:

Since there are no net external forces on the carts, momentum is conserved during the collision. Let’s define the 2.0\, \text {kg}2.0kg2, point, 0, start text, k, g, end text cart as object 111 and the 3.0\,\text {kg}3.0kg3, point, 0, start text, k, g, end text cart as object 222.

\begin{aligned}p_i &= p_f m_1 v_{1i} + m_2 v_{2i} &= (m_1 + m_2) v_{f} \end{aligned}  

p  

i

​  

m  

1

​  

v  

1i

​  

+m  

2

​  

v  

2i

​  

​  

=p  

f

​  

=(m  

1

​  

+m  

2

​  

)v  

f

​  

​  

Hint #22 / 3

Rearranging to solve for the final velocity of the carts gives:

\begin{aligned} v_f &= \dfrac{m_1 v_{1i} + m_2v_{2i}}{m_1+m_2}\\ \\ &= \dfrac{(2.0\, \text {kg})\left (5.0\,\dfrac{\text m}{\text s}\right ) + (3.0\, \text {kg})\left (-3.0\,\dfrac{\text m}{\text s}\right )}{2.0\,\text {kg}+3.0\,\text {kg}}\\ \\ &= 0.20\,\dfrac{\text m}{\text s}\end{aligned}  

v  

f

​  

​  

=  

m  

1

​  

+m  

2

​  

m  

1

​  

v  

1i

​  

+m  

2

​  

v  

2i

​  

​  

=  

2.0kg+3.0kg

(2.0kg)(5.0  

s

m

​  

)+(3.0kg)(−3.0  

s

m

​  

)

​  

=0.20  

s

m

​  

​  

Since speed is the magnitude of the velocity:

|v_{f}|=0.20\,\dfrac{\text m}{\text s}∣v  

f

​  

∣=0.20  

s

m

​  

vertical bar, v, start subscript, f, end subscript, vertical bar, equals, 0, point, 20, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction

Hint #33 / 3

The correct answer is 0.20\,\dfrac{\text m}{\text s}0.20  

s

m

​  

0, point, 20, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction.

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

they disappear if your absent from to long. i found out that the hard way.

step-by-step explanation:

brainliest lol i just need one more brainliest.