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2 years ago

If the average velocity during the athlete's walk back

1 answer:

4 0

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

From the question we are told

If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent is mathematically given as

T=\frac{d}{v}

Therefore

$T=\frac{d}{1.50}$

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

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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

2 years ago

Problem:

pav-90 [236]

Answer:

Total energy is constant

Explanation:

The laws of thermodynamics state that thermal energy (heat) is always transferred from a hot body (higher temperature) to a cold body (lower temperature).

This is because in a hot body, the molecules on average have more kinetic energy (they move faster), so by colliding with the molecules of the cold body, they transfer part of their energy to them. So, the temperature of the hot body decreases, while the temperature of the cold body increases.

This process ends when the two bodies reach the same temperature: we talk about thermal equilibrium.

In this problem therefore, this means that the thermal energy is transferred from the hot water to the cold water.

However, the law of conservation of energy states that the total energy of an isolated system is constant: therefore here, if we consider the hot water + cold water as an isolated system (no exchange of energy with the surroundings), this means that their total energy remains constant.

4 0

2 years ago

Which of the following could you do to increase the strength of an electromagnet?

Ann [662]

Wrap around a metal with wire instead of using wire alone.

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2 years ago

Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football

faltersainse [42]

Answer:

$a=159.32\ m/s^2$

Explanation:

It is given that,

Angular speed of the football spiral, $\omega=6.9\ rev/s=43.35\ rad/s$

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

$v=r\omega$

$v=0.085\times 43.35$

v = 3.68 m/s

The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(3.68)^2}{0.085}$

$a=159.32\ m/s^2$

So, the centripetal acceleration of the laces on the football is $159.32\ m/s^2$ . Hence, this is the required solution.

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2 years ago

Three sticks are arranged to form a right triangle. If the lengths of the three sticks are 0.47 m, 0.62 m and 0.78 m, what are t

Svetradugi [14.3K]

The angles in the triangle are 91 degrees, 53 degrees and 36 degrees respectively.

<h3>What is the cosine rule?</h3>

From the cosine rule we know that;

c^2 = a^2 + b^2 - 2abcosC

Since;

a = 0.47 m

b = 0.62 m

c = 0.78 m

Then;

(0.78)^2 = (0.47)^2 + (0.62)^2 - 2(0.47 * 0.62)cosC

0.61 = 0.22 + 0.38 - 0.58 cosC

0.61 - ( 0.22 + 0.38) = - 0.58 cosC

0.01 = - 0.58 cosC

C = cos-1(0.01/-0.58)

C = 91 degrees

Using the sine rule;

b/Sin B = c/Sin C

0.62/sinB = 0.78/sin 91

0.62/Sin B = 0.78

B = sin-1 (0.62//0.78)

B = 53 degrees

Angle A is obtained from the sum of angles in a triangle;

180 - (91 + 53)

A = 36 degrees

Learn more about triangle:brainly.com/question/2773823

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1 year ago