Ask question

Ask question

All categories

3 years ago

14

The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the t

wo boxes is _____

1 answer:

Nina [5.8K]3 years ago

4 0

Answer:

The speed of the boxes are 1 m/s.

Explanation:

Given that,

Mass of box = 1 kg

Mass of another box = 2 kg

Suppose 1 kg box moves with 3 m/s speed.

We need to calculate the speed of the boxes

Using formula of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

Where, u = initial velocity

v = final velocity

Put the value into the formula

$1\times3+2\times0=(1+2)v$

$v=\dfrac{3}{3}$

$v=1\ m/s$

Hence, The speed of the boxes are 1 m/s.

You might be interested in

In the context of depth perception, which of the following is a monocular cue?

olga55 [171]

Hello there.

<span>In the context of depth perception, which of the following is a monocular cue?

</span><span>(C) Convergence
</span>

8 0

3 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

Why is it important to understand the concept of density ?

N76 [4]

Density applies to many if not all aspects of life. With density you can explain why ice floats. You can explain why oxygen is on the earth, and not floating around in space( or being replaced by another gas). You can also explain why heat rises while cold air sinks.

8 0

2 years ago

When Missourians carve their initials into the bark of a tree, the damage leaves the tree open to ____________________.

sasho [114]

Answer:

In fact, carving letters into a tree probably won't hurt it. ... In general, the tree will compartmentalize the wound and it will heal over. The initials that remain visible are essentially scar tissue, permanent scar tissue.

Explanation:

Unfortunately, when carving into the trunk of a tree the blade of a knife often penetrates the outer bark and cuts into the inner bark. ... In cases that the phloem is damaged all the way around the trunk (in a ring for example), the tree will slowly and eventually starve to death.

add my s n a p

luke_raines19

5 0

2 years ago

A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th

Soloha48 [4]

Answer:

The velocity of mass 2m is $v_B = 0.67 m/s$

Explanation:

From the question w are told that

The mass of the billiard ball A is =m

The initial speed of the billiard ball A = $v_1$ =1 m/s

The mass of the billiard ball B is = 2 m

The initial speed of the billiard ball B = 0

Let the final speed of the billiard ball A = $v_A$

Let The finial speed of the billiard ball B = $v_B$

According to the law of conservation of Energy

$\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Substituting values

$\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Multiplying through by $\frac{1}{2}m$

$1 =v_A^2 + 2 v_B ^2 ---(1)$

According to the law of conservation of Momentum

$mv_1 + 2m(0) = mv_A + 2m v_B$

Substituting values

$m(1) = mv_A + 2mv_B$

Multiplying through by $m$

$1 = v_A + 2v_B ---(2)$

making $v_A$ subject of the equation 2

$v_A = 1 - 2v_B$

Substituting this into equation 1

$(1 -2v_B)^2 + 2v_B^2 = 1$

$1 - 4v_B + 4v_B^2 + 2v_B^2 =1$

$6v_B^2 -4v_B +1 =1$

$6v_B^2 -4v_B =0$

Multiplying through by $\frac{1}{v_B}$

$6v_B -4 = 0$

$v_B = \frac{4}{6}$

$v_B = 0.67 m/s$

4 0

3 years ago