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Cloud [144]
3 years ago
6

Betty (mass 40 kg), standing on slippery ice, catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Show that the

speed of Betty and her dog after the catch is about 0.8 m/s
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

v = 2.18m/s

Explanation:

In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.

Then you have:

Mv_{1o}+mv_{2o}=(M+m)v        (1)

M: mass Betty = 40kg

m: mass of the dog = 15kg

v1o: initial speed of Betty = 3.0m/s

v2o: initial speed of the dog = 0 m/s

v: speed of both Betty and her dog = ?

You solve the equation (1) for v:

v=\frac{Mv_{1o}+mv_{2o}}{M+m}=\frac{(40kg)(3.0m/s)+(15kg)(0m/s)}{40kg+15kg}\\\\v=2.18m/s

The speed fo both Betty and her dog is 2.18m/s

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A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i
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}.

3 0
3 years ago
An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o
jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

Substitute the values then we get

-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}

\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}

\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}

v=-\frac{20}{3}=-6.7 cm

Image distance of apple=-6.7 cm

Magnification=m=\frac{v}{u}=\frac{-\frac{20}{3}}{-20}

Magnification of apple=\frac{1}{3}=0.33

Hence, the magnification of apple=0.33

5 0
3 years ago
Find the equivalent resistance of this
frosja888 [35]

Answer:

Re=160ohm

Explanation:

Step#1

Rt=R1+R2 ( because both are in series)

Rt=(100+220 ) ohm

Rt=320 ohm

Step#2

Rt and R3 are parallel so,

Re= (Rt× R3) ÷ (Rt+R3)

Re= (320×320)÷( 320+320)

Re = 102,400÷ 640

Re=160ohm

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Which statement best explains the relationship between the electric force between two charged objects and the distance between t
spin [16.1K]
Unfortunately, the given statements are missing from the problem. However, we can still determine the relationship between the electric force between two objects and the distance between them. The formula for the electric force is given below:

F = (k*Q1*Q2)/d^2

k is a constant, while Q1 and Q2 are the respective charges of the objects. F is force, while d is distance.

As seen in the formula, we can see that the electric force F is inversely proportional to the square of the distance between the two objects.
3 0
3 years ago
A car is travelling at 18m/s accelerates ti 30m/s in 3seconds. what's the acceleration of the car​
Luden [163]

Answer: a = 4 m/s²

Explanation:

a = Δv/t = (30 - 18) / 3 = 4 m/s²

3 0
3 years ago
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