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Ilia_Sergeevich [38]

3 years ago

11

Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8

kg, 31.5 kg, and 25.9 kg, respectively.
What is the new velocity of the sled if Kid A jumps off vertically from the sled? (Kid A velocity = 0)Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8 kg, 31.5 kg, and 25.9 kg, respectively.

What is the new velocity of the sled if Kid A jumps off vertically from the sled? (Kid A velocity = 0)

1 answer:

Kamila [148]3 years ago

4 0

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

$V2=\frac {M1v1}{m2}$

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

$V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s$

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Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

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Describe a change caused by kinetic energy as well as a change that involves potential energy

kykrilka [37]

Kinetic energy is formed when the object is in motion.
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One of the example is Corn flour factory.

Corn turned into flour by a windmill that moved by the waterfall. Movement of the mill is relative to the power given by waterfall (potential energy) and the spinning crushes the corn into flour (kinetic energy)

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1. When you have different masses for each sphere, how does the force that the larger mass sphere exerts on the smaller mass sph

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1) The forces are equal (Newton's third law of motion)

2) The force between the spheres will quadruple

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Explanation:

1)

In this part of the problem, we want to compare the gravitational force exerted by the larger mass sphere on the smaller mass sphere to the force exerted by the smaller mass sphere to the larger mass sphere.

We can do this by using Newton's third law of motion, which states that:

<em>"When an object A exerts a force (called </em><em>action</em><em>) on an object B, then object B exerts an equal and opposite force (called </em><em>reaction</em><em>) on object A"</em>

In this problem, we can identify the larger mass sphere as object A and the smaller mass sphere as object B. This law tells us that the two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the larger mass sphere on the smaller mass sphere is equal to the force exerted by the smaller mass sphere to the larger mass sphere.

2)

The magnitude of the gravitational force between the two spheres is given by

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

$m_1, m_2$ are the masses of the two spheres

r is the separation between the two spheres

In this problem, we are asked to find what happens when the distance between the spheres is halved, therefore when the new distance is

$r'=\frac{r}{2}$

Substituting into the equation, we find

$F'=G\frac{m_1 m_2}{r'^2}=G\frac{m_1 m_2}{(r/2)^2}=4(\frac{Gm_1 m_2}{r^2})=4F$

So, the force between the two spheres will quadruple.

3)

We can give an estimate for the gravitational force exerted by your notebook on you.

As we said, the magnitude of the gravitational force is

$F=G\frac{m_1 m_2}{r^2}$

Where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

Let's estimate the following:

$m_1 = 60 kg$ is your mass

$m_2 = 2 kg$ is the mass of the notebook

$r=1 m$ , assuming the notebook is at 1 metre from you

Substituting,

$F=(6.67\cdot 10^{-11})\frac{(60)(2)}{1^2}=8.0\cdot 10^{-9} N$

We see that this force has an extremely small value: therefore, it is almost negligible in daily life, where other much stronger forces act on you.

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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