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3 years ago

List five different institutional sources of contracts that you know

1 answer:

6 0

The contract of citizenship, the contract of merriage, the contract of loan, the contract of power, and the contract of health care. All these are examples of a contract on institutional level

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What acceleration occurs if the friction force in #4 is now 4N?

anastassius [24]

Answer:

you a b1tc h

Explanation:

you're not going anywhere in life. mcdonalds hiring $16 an hour where i live.

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2 years ago

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How is magnetism important to our evolution?

sweet [91]

If it helps Mark Brainliest.. :)

natural magnetism of the Earth derives from its iron core. This not only provides a useful direction finder for compasses, but actually protects life on Earth by deflecting charged particles in space. The "magnetosphere" is a large region that surrounds the Earth as it moves in its orbit around the Sun. It consists of charged ions that are prevented from directly striking the surface, where they could injure living organisms and harm the environment.When solar eruptions on the Sun increase the flow of charged particles, industries such as power transmission and communication can be still be affected despite the magnetic field

4 0

2 years ago

Magnetism is due to the movement of ______.

Ierofanga [76]

Answer:

electrons

Explanation:

Brainliest

5 0

2 years ago

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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

A ball of mass 2 kg is kept on the hill of height 3 km. Calculate the potential energy possessed by it ?

zysi [14]

We know that -

$P.E=m*g*h$

Where,

$m = mass$

$g = acceleration$ $due$ $to$ $gravity$

$h=height$

First we convert height into meters.

1 km = 1000 meters

3 km = 1000 * 3 meters = 3000 meters

So, putting the values in the above formula, and by taking 'g' = 9.8 m/s², we get-

$P.E.= 2*3000*9.8$

$P.E.= 58800$ $Joules$

$P.E.= 58.8$ $kJ$

7 0

3 years ago

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