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Brut [27]
3 years ago
10

A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Upper P 89​, the

89 th percentile. This is the temperature reading separating the bottom 89 % from the top 11 %.

Physics
1 answer:
andrew11 [14]3 years ago
8 0

Answer:

X_{1}=1.23

Explanation:

Given :

μ = 0

σ = 1

For 89th percentile

using invariant norm ( area, μ, σ )

       = inv. norm ( 0.89, 0, 1 )

       = 1.23

or

P ( x > X_{1} ) = 0.89

P\left ( \frac{X-N}{\sigma } >\frac{X_{1}-0}{1}\right )=0.89

P\left ( Z>Z_{1} )=0.89

Now using Normal table, Z = 1.23

Therefore, \frac{X_{1}-0}{1}=1.23

X_{1}=1.23

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grin007 [14]
<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

Refractive index of medium = speed of light in vacuum / speed of light in medium  

So

RI = 1/0.6 = 5/3 or 1.66

3 0
3 years ago
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Answer:

Explanation:

radius of hoop and the radius of disk is same = R

Let the mass of hoop is M and the mass of disk is M'.

As they reach the bottom of teh surface in same time so they travel equal distance thus, they have same acceleration.

The acceleration is given by

a=\frac{gSin\theta }{1+\frac{I}{MR^{2}}}

As the acceleration is same so that the moment of inertia is also same.

Moment of inertia of disk = moment of inertia of hoop

1/2 x mass of disk x R² =  mass of hoop x R²

So, mass of disk = 2 x mass of hoop

Option (c) is correct.

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3 years ago
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MrRa [10]

Answer:

Diorite, Gabbro, and Granite

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A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05
7nadin3 [17]

Answer:

W_{drag} = 4.223\,J

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}

The work done on the ball due to drag is:

W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})

W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})

W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

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7 0
3 years ago
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A student's calculation for the work done on an object is shown.
MissTica

The statement that describes the error in the work is that the distance must be converted to meters (m).

<h3>FORMULA FOR WORK:</h3>

Work can be calculated by using the following formula:

W = F × d

Where;

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According to this question, the force is given as 140N and the distance is given as 30cm. The force is calculated as follows:

F = 140N × 30cm = 4200J

This calculation is erroneous because the unit of distance should be converted from cm to meters.

Learn more about work done at: brainly.com/question/3902440

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