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Wittaler [7]
3 years ago
5

What is the kinetic energy of a 50-kg child running to catch the school bus at

Physics
2 answers:
Vinvika [58]3 years ago
6 0

Answer:

Option C

100 J

Explanation:

Kinetic energy, KE is given by

KE=0.5mv^{2} where m is the mass and v is the velocity

Substituting 50 Kg for mass, m and 2 m/s for velocity v then we obtain

KE=0.5*50*2^{2}=100 J

Therefore, the child's kinetic energy is equivalent to 100 J

Zepler [3.9K]3 years ago
5 0

Answer:100J

Explanation:

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3 years ago
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Writing a 4 to 5 sentence paragraph, identify YOUR stance on whether more Nuclear Reactors should be built here in Michigan. (1s
prisoha [69]
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3 years ago
Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orb
murzikaleks [220]

Answer:

  L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

         L = r xp

         L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

        x_{cm} = \frac{1}{M} \sum  x_{i} m_{i}i

        x_{cm} = \frac{1}{m+m} ( 0 + l m)

        x_{cm} = \frac{1}{2m}  lm

        x_{cm} = \frac{1}{2}

        x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

          L = 2  6.05  74.3  5.65

          L = 5076.5 kg m² / s

4 0
3 years ago
Read 2 more answers
The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of
Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is  

 \theta _d  =0.19 ^o

Explanation:

From the question we are told that

     The  refractive index of the transparent acrylic plastic for blue light is  n_F  =  1.497

     The  wavelength of the blue light is F  =  486.1 nm  =  486.1 *10^{-9} \ m

    The  refractive index of the transparent acrylic plastic for red light is  n_C  =  1.488

       The  wavelength of the red light is C =  656.3 nm  = 656.3 *10^{-9} \  m

    The incidence angle is  i  =  45^o

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       r_F =  sin ^{-1}[\frac{sin(i) *  n_a }{n_F} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_F =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.497} ]

      r_F  =  28.18^o

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       r_C =  sin ^{-1}[\frac{sin(i) *  n_a }{n_C} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_C =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.488} ]

      r_F  =  28.37^o

The angle between the blue beam and the red beam in the acrylic block

     \theta _d  =  r_C  - r_F

substituting values

       \theta _d  = 28.37 -  28.18

       \theta _d  =0.19 ^o

 

4 0
3 years ago
suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area vapour already present in su
gulaghasi [49]
That's two different things it depends on:

-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.

Here's what I have in mind for an experiment to show those two dependencies:

-- a closed box with a wall down the middle, separating it into two closed sections;

-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.

-- a tiny fan that blows air through a tube into the hole in one outer wall.

<u>Experiment A:</u>

-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================

<u>Experiment B:</u>

-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section.  Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
6 0
3 years ago
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