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3 years ago

What are higher-than-usual tides that form when the suns and moons gravitational pulls work together

Physics

1 answer:

Soloha48 [4]3 years ago

8 0

These are called neap tides!

Hope this helps :)

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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

Freud believes that defense mechanisms are reactions people have to protect themselves from uncomfortable feelings. the article

makkiz [27]

Answer:

A: Anxious

Explanation:

got a 100% of Edge

6 0

2 years ago

astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o

devlian [24]

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ $T=2 \pi\sqrt{\frac{m}{k} }$

or,

⇒ $m=\frac{k T^2}{4 \pi^2}$

On substituting the values, we get

⇒ $=\frac{280\times (3)^2}{4 \pi^2}$

⇒ $=\frac{280\times 9}{4\times (3.14)^2}$

⇒ $=68.83 \ kg$

(b)

The speed of the string will be:

⇒ $\frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2$

then,

⇒ $v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }$

On substituting the values, we get

⇒ $=\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }$

⇒ $=\sqrt{\frac{280(0.16-0.04)}{63.83} }$

⇒ $=\sqrt{\frac{280\times 0.12}{63.83} }$

⇒ $=0.725 \ m/s$

4 0

3 years ago

In which of the following situations would convection currents most likely occur?

poizon [28]

Answer: Option (B) is the correct answer.

Explanation:

Convection is defined as a process in which a fluid (liquid or gas) is heated and the cooler liquid (more dense) settles or sinks at the bottom whereas a hotter material (less dense) will rises to the top.

For example, ice being less dense (cooler material) will float in a lake whereas liquid water being more dense will sink at the bottom.

Thus, we can conclude that situation in which convection currents most likely occur is within the water in a lake as ice forms on the top.

4 0

3 years ago

Read 2 more answers

Block 1 (mass 2.00 kg) is moving rightward at 10.0 m/s and block 2 (mass 5.00 kg) is moving rightward at 3.00 m/s. The surface i

DaniilM [7]

Answer:

a) 0.25m

b) 5 m/s

Explanation:

When the spring is compressed both boxes are moving with the same velocity, so applying the principle of linear momentum conservation:

$m1*v_{o1}+m2*v_{o2}=(m1+m2)*v\\v=5m/s$

Now applying the principle of energy conservation:

$K1+K2+U_{g1}-U_e=Kf+U_{g2}\\K1+0-U_e=K2+0\\U_e=K1+K2-kf\\\frac{1}{2}*k*x^2+=\frac{1}{2}*m1*v1^2+\frac{1}{2}*m1*v1^2-\frac{1}{2}*(m1+m2)*v^2\\\\x=\sqrt{\frac{2.00kg*(10m/s)^2+5.00kg*(3.00m/s)^2-7.00kg*(5m/s)^2}{1120N/m}}\\x=0.25m$

We got that the maximum compression is 0.25m.

5 0

2 years ago