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Nataly_w [17]
2 years ago
8

1. Which of the following would be a testable hypothesis?

Physics
1 answer:
Naddik [55]2 years ago
7 0
B dropping a ball
C tentative and testable
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A young man walks daily through a gridded city section to visit his girlfriend, who lives m blocks East and nblocks North of whe
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Answer:

The man ate eggs.

Explanation:

He should brush his teeth before seeing his girlfriend.

4 0
3 years ago
All of the following are functions of the skeletal system EXCEPT:
sasho [114]
I'm pretty sure the answer would be D
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3 years ago
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What is the moment of inertia of the object starting from rest if it has a final velocity of 5.9 m/s? Express the moment of iner
Bogdan [553]

Answer:

The moment of inertia is I = 0.126*R^2*M

Explanation:

We can calculate the moment of inertia of an object that starts from rest and has a final velocity using the energy conservation equation, as follows:

Ek1 + Ep1 = Ek2 + Ep2, where

Ek1 = kinetic energy of the object before to roll down

Ep1 = potential energy of the object

Ek2 = kinetic energy when the object comes down

Ep2 = potential energy of the object at the bottom

We have the follow:

Ek1 = 0

Ep1 = M*g*h

Ek2 = ((I*w)/2) + ((M*v^2)/2)

Ep2 = 0

Replacing values:

0 + M*g*h = ((I*w)/2) + ((M*v^2)/2) + 0

where:

M = mass of the object

g = gravitational acceleration

I = moment of the inertia

w = angular velocity = v/R

h = height

M*g*h = ((1/2) * I * (v^2/R^2)) + ((M*v^2)/2)

M*9.8*2 = (I*(5.9^2)/(2*R^2)) + ((5.9^2 * M)/2)

19.6 * M = ((17.4*I)/R^2) + 17.4*M

Clearing I, we have:

I = 0.126*R^2*M

5 0
3 years ago
Light of wavelength 520 nm is used to illuminate normally two glass plates 21.1 cm in length that touch at one end and are separ
Oksi-84 [34.3K]

Answer:

The number is  Z = 216 \ fringes

Explanation:

From the question we are told that

      The wavelength is  \lambda  =  520 \ nm =  520 *10^{-9} \ m

       The length of the glass plates is y  = 21.1cm = 0.211 \ m

      The distance between the plates (radius of wire ) =  d =  0.028 mm  =  2.8 *10^{-5} \  m

   Generally the condition for constructive  interference in a film is mathematically represented as

            2 *  t  = [m +  \frac{1}{2}  ]\lambda

Where  t is the thickness of the separation between the glass i.e  

    t  = 0 at the edge where the glasses are touching each other and  

     t =  2d at the edge where the glasses are separated by the wire  

   m is the order of the fringe it starts from  0, 1 , 2 ...

So  

       2 *  2 * d   = [m +  \frac{1}{2}  ] 520 *10^{-9}

=>   2 *  2 *   (2.8 *10^{-5}) = [m +  \frac{1}{2}  ] 520 *10^{-9}

=>    

       m = 215

given that we start counting m from zero

   it means that the number of  bright fringes that would appear is

         Z =  m + 1

=>    Z =  215 +1

=>     Z = 216 \ fringes

3 0
3 years ago
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4 0
3 years ago
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