Three point mass particles are located in a plane: a. 3.77 kg located at the origin
b. 6.7106 kg at [(5.72 cm),(11.44 cm)]
c. 2.46181 kg at [(16.7024 cm),(0 cm)].

How far is the center of mass of the three particles from the origin? Answer in units of cm

Answers

Answer 1

Answer:

The distance of the center of mass of the three particles from the origin is 6.1428 cm and 5.9316 cm.

Calculation of the distance:

Since

m1 = 3.77 kg (0, 0 )

m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)

m3 = 2.46181 kg (16.7024 cm, 0 cm )

Now here we assume x and y be the coordinates with respect to the centre of mass.

So,

We know that

$x = (m_1x_1+m_2x_2+m_3x_3)/(m_1+m_2+m_3)\n\n = (3.77* 0 + 6.7106 * 5.72 + 2.46181 * 16.7024)/(3.77 + 6.7106 + 2.46181)$

= 6.1428 cm

Now

$y = (m_1y_1+m_2y_2+m_3y_3)/(m_1+m_2+m_3)\n\n = (3.77* 0 + 6.7106 * 11.44 + 2.46181 * 0)/(3.77 + 6.7106 + 2.46181)$

= 5.9316 cm

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Answer 2

Answer:

Answer:

Explanation:

m1 = 3.77 kg (0, 0 )

m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)

m3 = 2.46181 kg (16.7024 cm, 0 cm )

Let x and y be the coordinates of centre of mass.

$x = (m_(1)x_(1)+ m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))$

$x = (3.77* 0+ 6.7106* 5.72 + 2.46181* 16.7024)/(3.77+6.7106+2.46181)$

x = 6.1428 cm

$y = (m_(1)y_(1)+ m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))$

$y= (3.77* 0+ 6.7106* 11.44 + 2.46181* 0)/(3.77+6.7106+2.46181)$

y = 5.9316 cm

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If a material has an index of refraction of 1.61, Determine the speed of light through this medium

Answers

Answer:

1.86 x 10^8 m/s

Explanation:

n = 1.61

The formula for the refractive index is given by

n = speed of light in vacuum / speed of light in material

n = c / v

v = c / n

v = (3 x 10^8) / 1.61

v = 1.86 x 10^8 m/s

Final answer:

The speed of light in a material with an index of refraction of 1.61 is calculated as approximately 1.86 * 10^8 m/s, using the equation v = c/n where c is the speed of light in vacuum and n is the index of refraction.

Explanation:

The speed of light in a given material can be calculated using the index of refraction of the material, as defined by the equation n = c/v, where n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the material.

Given that the index of refraction for the material in question is 1.61, and the speed of light in vacuum, c = 3.00 * 10^8 m/s, the speed of light v in this medium would therefore be calculated by rearranging the equation to v = c/n.

By substituting the given values into the equation, v = 3.00 * 10^8 m/s / 1.61, we find that the speed of light in the material is approximately 1.86 * 10^8 m/s.

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The position of a particle as a function of time is given by x = (2.0 m/s)t + (-3.0 m/s2)t2. (a) plot x-versus-t for time from t = 0 to t = 1.0 s. (do this on paper. your instructor may ask you to turn in this plot.) this answer has not been graded yet. (b) find the average velocity of the particle from t = 0.45 s to t = 0.55 s. m/s (c) find the average velocity from t = 0.49 s to t = 0.51 s.

Answers

Part a)

Equation of position with time is given as

$x = (2.0 m/s)t + (-3.0 m/s2)t^2$

since this equation is a quadratic equation

so it will be a parabolic graph between t = 0 to t = 1

part b)

at t = 0.45 s

$x = 2* 0.45 - 3 * 0.45^2$

$x_1 = 0.2925 m$

at t = 0.55 s

$x = 2* 0.55 - 3*0.55^2$

$x_2 = 0.1925$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.1925 - 0.2925 = -0.1 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.1}{0.1) = -1 m/s$

part c)

at t = 0.49 s

$x = 2* 0.49 - 3 * 0.49^2$

$x_1 = 0.2597 m$

at t = 0.51 s

$x = 2* 0.51 - 3*0.51^2$

$x_2 = 0.2397 m$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.2397 - 0.2597 = -0.02 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.02}{0.02) = -1 m/s$

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 76.2-kg water-skier has an initial speed of 5.0 m/s. Later, the speed increases to 10.4 m/s. Determine the work done by the net external force acting on the skier.

Answers

Answer:

Work done will be equal to 3186.396 J

Explanation:

We have mass m = 76.2 kg

Initial velocity u = 5 m/sec

Final velocity v = 10.4 m/sec

We have to find the work done

From work energy theorem work done is equal to change in kinetic energy

$w=(1)/(2)mv^2-(1)/(2)mu^2$

$w=(1)/(2)* 76.2* 10.4^2-(1)/(2)* 76.2* 5^2$

w = 3168.396 J

So work done will be equal to 3186.396 J

Classical mechanics is an extremely well tested model. Hundreds of years worth of experiments, as well as most feats of engineering, have verified its validity. If special relativity gave very different predictions than classical physics in everyday situations, it would be directly contradicted by this mountain of evidence. In this problem, you will see how some of the usual laws of classical mechanics can be obtained from special relativity by simply assuming that the speeds involved are small compared to the speed of light.Two of the most surprising results of special relativity are time dilation and length contraction, namely, that measured intervals in time and space are not absolute quantities but instead appear differently to different observers. The equations for time dilation and length contraction can be written t=?t0 and l=l0/?, where?=11?u2c2?.Part AFind the first two terms of the binomial expansion for ?.Express your answer in terms of u and c.Hints? = 1+12(uc)2 … SubmitMy AnswersGive UpCorrectYou can see that ??1 if u?c, as is the case in most situations. If you set ?=1 in the equations for time dilation and length contraction you recover the equations of classical physics, which state essentially that there is no time dilation or length contraction. Therefore, we don't see any appreciable length contraction or time dilation in everyday life.Part BConsider a case involving a speed that is fast compared to those encountered in our everyday life: a spy plane moving at 1500m/s. Find the deviation from classical physics (??1) that relativity predicts at this speed. Use only the first two terms of the binomial expansion, as your calculator may not be able to handle the necessary number of digits otherwise.Express your answer to four significant figures.??1 = 1.250×10?11SubmitMy AnswersGive UpCorrectIf you lived for 70 years in such a spy plane moving at 1500m/s, this would amount to about 28ms of cumulative time difference between you and people who lived at rest relative to the earth when you finally landed. Thus, it is not surprising that relativistic effects are not observed in everyday life, or even at the fringes of everyday life. By using atomic clocks, which can measure time accurately to one part in 1013 or better, the time dilation at the normal speed for an airliner has been verified.Part CNow, consider the relativistic velocity addition formula:speed=v+u1+vuc2.If v=u=0.01c=1% of c, what is the relativistic sum of the two speeds?Express your answer as a percentage of the speed of light to five significant figures.

Answers

Answer:

The Answer is 0.019998c

Explanation:

Please see the attached Picture for the answer.

A golf pro swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00340 s. After the collision, the ball leaves the club at a speed of 46.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?

Answers

To solve this problem it is necessary to apply the concepts related to Newton's second law and the equations of motion description for acceleration.

From the perspective of acceleration we have to describe it as

$a = (\Delta v)/(\Delta t)$

Where,

$\Delta v$ = Velocity

$\Delta t$ = time

At the same time by the Newton's second law we have that

F = ma

Where,

m = mass

a = Acceleration

Replacing the value of acceleration we have

$F = m ((\Delta v)/(\Delta t))$

Our values are given as,

$m = 55*10^(-3)Kg$

$v = 46m/s$

$t = 0.00340s$

Replacing we have,

$F = m ((\Delta v)/(\Delta t))$

$F = (55*10^(-3))((46)/(0.00340))$

$F = 744.11N$

Therefore the magnitude of the average force exerted on the ball by the club is 744.11N

Sheba is texting and driving. She does not see the baby chinchilasitting in the road! If she was traveling at 35 m/s, and at best her
car can accelerate at -3.5 m/p/s, how far would it take her to stop?