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tamaranim1 [39]
3 years ago
13

Un cuerpo gira con movimiento circular uniforme un ángulo total de 1080° en 20

Physics
1 answer:
LekaFEV [45]3 years ago
7 0

Velocidad angular = (angulo total) / (tiempo total)

Velocidad angular = (1080 grados) / (20 segundos)

Velocidad angular = (1080/20) g/s

Velocidad angular = 54 g/s

Pero 180 grados = π radianes

V.A. = (54 g/s) x (π rad / 180 g)

V.A. = (54π gr-rad / 180 seg-gr)

<em>V.A. = 0.3π rad/seg</em>

V.A. = aproximadamente  0.942 rad/seg

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Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68
creativ13 [48]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is   B= 4.2 *10^ {-6}T , the direction is into the page

Explanation:

From the question we are told that

        The current  is i = 12.0 A

        The radius of arc  bc is r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m

        The radius  of arc da is r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m

        The length of segment cd and ab is = l = 10cm = \frac{10}{100} = 0.10 m

The objective of the solution is to obtain the magnetic field

    Generally magnetic due to the current flowing in the arc is mathematically represented as

             B = \frac{\mu_o I}{4 \pi r}

 Here I is the current

         \mu_o is the permeability of free space with a value of 4\pi *10^{-7}T \cdot m/A

            r is the distance

Considering Arc da

         B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta

Where \theta is the angle the arc da makes with the center  from the diagram its value is  \theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad

     Now substituting values into formula for magnetic field for da

                    B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]

                           = \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]

                   B_{da}= 12.56*10^{-6} T

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

             B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta

Substituting value

          B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]

                B_{bc}= 8.37*10^{-6}T

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of  the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

     Then the net magnetic field would be

                 B = B_{da} - B{bc}

                     = 12.56*10^{-6} - 8.37*10^{-6}

                     = 4.2 *10^ {-6}T

       Since B_{da} > B_{bc} then the direction of the net charge would be into the page

 

3 0
4 years ago
Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a
Sati [7]

Ans;

6.25m²/Kd

Explanation: see attached file

5 0
3 years ago
Dr. John Paul Stapp was U.S. Air Force officer whostudied the effects of extreme deceleration on thehuman body. On December 10,
Ira Lisetskai [31]

Answer:

(a) a = 56.4 m/s², his acceleration a, in multiples of gravity g, is 5.76 g

(b) a = -201.43 m/s², his deceleration -a, in multiples of gravity g, is -20.56 g

Explanation:

(a)

When moving upwards, the initial velocity, u = 0 (he accelerated from rest)

When moving upwards, the final velocity, v = 282 m/s

time of  motion during this acceleration, t = 5 s

His acceleration is calculated as;

v = u + at

282 = 0 + 5a

a = 282 / 5

a = 56.4 m/s²

Ratio of his acceleration, a to gravity, g = a/g = 56.4 / 9.8 = 5.76

a = 5.76 g

(b)

When moving downwards, the initial velocity, u = 282 m/s

When moving downwards, the final velocity, v = 0 (he was brought to rest)

time of  motion during this deceleration, t = 1.4 s

His deceleration is calculated as;

v = u + at

0 = 282 + 1.4a

1.4a = -282

a = -282 / 1.4

a = -201.43 m/s²

Ratio of his deceleration, -a to gravity, g = -a/g = 201.43 / 9.8 = 20.56

a = -20.56 g

3 0
3 years ago
What are 3 criteria for acceleration?
Temka [501]

I don't think you mean 'criteria'.  I think you mean three occurrences or
observations that indicate the presence of acceleration.

They are:

-- an object is speeding up
-- an object is slowing down
-- the direction of an object's motion is changing .

Any one of these changes is acceleration.

There's a single term that covers them all.  It is "change in velocity".


3 0
3 years ago
I NEED HELP ASAP
LiRa [457]

The object is called a meteor because it is producing Streak of light  and has not yet struck earth.

<h3><u>Explanation:</u></h3>

A meteoroid is a celestial object which is very smaller than an asteroid. These objects are produced as a collision impact from mars or moon and float freely in space without any specific orbit. When they come inside the Earth's gravitational field, they are attracted by the Earth's gravity to Earth's crust. These objects in Earth's atmosphere are called meteors. As they travel through Earth's atmosphere, they do face a huge friction from Earth's atmosphere which let them burn and that is visible as the tail of the meteor.

Most of them are so small that they are burnt away in the atmosphere. But some are bigger and they reach the Earth's surface and are called as meteorites.

4 0
3 years ago
Read 2 more answers
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