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Nimfa-mama [501]
3 years ago
12

List five different institutional sources of contracts that you know

Physics
1 answer:
dmitriy555 [2]3 years ago
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
You might be interested in
A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b
Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=\frac{1}{2} kx^{2} +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒\frac{1}{2} kx^{2} +PE1=PE2

⇒PE2-PE1=\frac{1}{2} kx^{2}

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

⇒mgh=\frac{1}{2} kx^{2}

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴2100*9.8*h=\frac{1}{2}*2200*0.11^{2}

⇒20580*h=13.31

⇒h=\frac{13.31}{20580}

⇒h=0.0006467m=6.5e-4

7 0
4 years ago
Find the volume of a sphere of radius 10 mm.
vredina [299]

Answer:

Explanation: This is done using the equation:

\frac{4}{3} π R^{3}

Because the Radius is a know value. We have the following.

\frac{4}{3} π (10mm)^{3}

Which is:

4188.7902 mm

5 0
3 years ago
A diffraction grating has 500 slits/mm. What is the longest wavelength of light for which there will be a third-order maximum?
Alexxandr [17]

Answer:

The longest wavelength of light  is 666.7 nm

Explanation:

The general form of the grating equation is

mλ = d(sinθi + sinθr)

where;

m is third-order maximum = 3

λ is the wavelength,

d is the slit spacing (m/slit)

θi  is the incident angle

θr is the diffracted angle

Note: at longest wavelength, sinθi + sinθr = 1

λ = d/m

d = 1/500 slits/mm

λ = 1 mm/(500 *3) = 1mm/1500 = 666.7 X 10⁻⁶ mm = 666.7 nm

Therefore, the longest wavelength of light  is 666.7 nm

8 0
3 years ago
4) A satellite, mass m, is in circular orbit (radius r) around the earth, which has mass ME and radius Re. The value of r is lar
defon
<h2>Answers:</h2>

(a) The kinetic energy of a body is that energy it possesses due to its movement and is defined as:

K=\frac{1}{2}m{V}{2}     (1)

Where m is the mass of the body and V its velocity.

In this specific case of the satellite, its kinetic energy K_m taking into account its mass m is:

K_{m}=\frac{1}{2}m{V}^{2}     (2)

On the other hand, the velocity of a satellite describing a circular orbit is constant and defined by the following expression:

V=\sqrt{G\frac{ME}{r}}     (3)

Where G is the gravity constant, ME the mass of the Earth and r the radius of the orbit <u>(measured from the center of the Earth to the satellite). </u>

Now, if we substitute the value of V from equation (3) on equation (2), we will have the final expression of the kinetic energy of this satellite:

K_{m}=\frac{1}{2}m{\sqrt{G\frac{ME}{r}}}^{2}     (4)

Finally:

K_{m}=\frac{1}{2}Gm\frac{ME}{r}     (5)  >>>>This is the kinetic energy of the satellite

(b) According to Kepler’s 2nd Law applied in the case of a circular orbit, its Period T is defined as:

T=2\pi\sqrt{\frac{r^{3}}{\mu}}     (6)

Where \mu is a constant and is equal to GME. So, this equation in these terms is written as:

T=2\pi\sqrt{\frac{r^{3}}{GME}}     (7)

As we can see, <u>the Period of the orbit does not depend on the mass of the satellite </u>m, it depends on the mass of the greater body (the Earth in this case) ME, the radius of the orbit and the gravity constant.

(c) The gravitational force described by the law of gravity is a central force and therefore is <u>a conservative force</u>. This means:

1. The work performed by a gravitational force to move a body from a position A to a position B <u>only depends on these positions and not on the path followed to get from A to B. </u>

2. When the path that the body follows between A and B is a c<u>losed path or cycle</u> (as this case with a <u>circular orbit</u>), <u>the gravitational work is null or zero</u>.

<h2>This is because the gravity force that maintains an object in circular motion is a centripetal force, that is, <u>it always acts perpendicular to the movement</u>. </h2>

Then, in the case of the satellite orbiting the Earth in a circular orbit, its movement will always be perpendicular to the gravity force that attracts it to the planet, at each point of its path.

(d)  The total Mechanical Energy E of a body is the sum of its Kinetic Energy K and its Potential Energy P:

E=K+P     (8)

But in this specific case of the circular orbit, its kinetic energy will be expresses as calculated in the first answer (equation 5):

K_{m}=\frac{1}{2}Gm\frac{ME}{r}     (5)

And its potential energy due to the Earth gravitational field as:

P_{m}=-G\frac{mME}{r}     (9)

This energy is negative by definition.

So, the total mechanical energy of the orbit, also called the Orbital Energy is:

E=\frac{1}{2}Gm\frac{ME}{r}+(- G\frac{mME}{r})      (10)

Solving equation (10) we finally have the Orbital Energy:

E=-\frac{1}{2}mME\frac{G}{r}     (11)

At this point, it is necessary to clarify that a satellite (or any other celestial body) orbiting another massive body, can describe one of these types of orbits depending on its Orbital Total Mechanical Energy E:

-When E=0:

We are talking about an <u>open orbit</u> in which the satellite escapes from the attraction of the planet's gravitational field. The shape of its trajectory is a parabola, fulfilling the following condition:

K_{m}=-P_{m}

Such is the case of some comets in the solar system.

-When E>0:

We are also talking about <u>open orbits</u>, which are hyperbolic, being K_{m}>P_{m}

<h2>-When E: >>>><u>This case</u></h2>

We are talking about <u>closed orbits</u>, that is, the satellite will always be "linked" to the gravitational field of the planet and will describe an orbit that periodically repeats with a shape determined by the relationship between its kinetic and potential energy, as follows:

-Elliptical orbit: Although E is constant, K_m and P_m are changing along the trajectory .

-Circular orbit: When at all times both the kinetic energy K_m and the potential P_m remain constant, resulting in a total mechanical energy E as the one obtained in this exercise. This means that the speed is constant too and <u>is the explanation of why this Energy has a negative sign. </u>

3 0
3 years ago
A 12 kg bowling ball would require what force to accelerate it down an alley at a rate of 2.5 m/s ^ 2
alisha [4.7K]

Answer:

hi

Explanation:

hiijjjjjjjjjjjjjjj

8 0
3 years ago
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