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Alisiya [41]
3 years ago
12

A thin wire of length (2 m) and (1 mm2 ) cross-section area is clamped horizontally between two walls, a weight of (10 kg) is hu

ng at the middle of the wire depressed it (2 cm). Find young's modulus?
Physics
1 answer:
allsm [11]3 years ago
4 0

Answer:

Young modulus  = 9.8 × 10⁹ N/m²

Explanation:

From the information given:

Stress = F/A

Stress = (10 × 9.8) / 0.001²

Stress = 9.8× 10⁷ N/m²

Strain = increase in length / initial length of wire

Strain = 0.02/ 2

Strain = 0.01

Now;

The Young modulus (Y)= stress/strain

Young modulus  = (9.8 × 10⁷ N/m²) /  0.01

Young modulus  = 9.8 × 10⁹ N/m²

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Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.
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The velocity would be the same for all ramps.
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The seismic activity density of a region is the ratio of the number of earthquakes during a given time span to the land area aff
Natalija [7]

Answer:

0.0059

Explanation:

According to the question the seismic activity density is given by

\text{Seismic activity density}=\frac{\text{Number of Earthquakes over a given time span}}{\text{The land area affected}}

Here,

Number of Earthquakes over a given time span = 424

The land area affected = 71300 mi²

So,

\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059

The seismic activity density is 0.0059

8 0
3 years ago
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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
4 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
3 years ago
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