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3 years ago

5

Professor Frank Nabarro insists that all senior physics majors take his notorious physics aptitude test. The test is so tough th

at anyone not going on to a career in physics has no hope of passing, whereas 35% of the seniors who do go on to a career in physics still fail the test. Further, 70% of all senior physics majors in fact go on to a career in physics. Assuming that you fail the test, what is the probability that you will not go on to a career in physics? (Round your answer to four decimal places.)

1 answer:

yulyashka [42]3 years ago

8 0

Answer:

0.5504

Explanation:

We can use Bay's theorem to solve this question

P( No physics career/fail) = P(No Physics career and fail)/P(Fail)

= P( Fail/ No physics career)×P( no physics career)/P( fail)

= $\frac{0.3\times1}{(0.7\times0.35)+(0.3\times1)}$

=0.5504

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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

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Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

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3 years ago

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The change in kinetic energy is $\Delta K = 3Fd$

Explanation:

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$W=K_f -K_i= \Delta K$

where :

W is the work done on the object

$K_f$ is the final kinetic energy of the object

$K_i$ is the initial kinetic energy

Also, the work done on an object is (assuming that the force is applied parallel to the motion of the object):

$W=F\Delta x$

where

F is the magnitude of the force

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In this problem, the force acting on the object is

F

While the displacement is the horizontal distance travelled, so

$\Delta x = 3d$

Therefore, the work done is

$W=(F)(3d)=3Fd$

And so the change in kinetic energy is

$\Delta K = 3Fd$

Learn more about work and kinetic energy:

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