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Trava [24]
3 years ago
11

A man does 500 J of work pushing a car a distance of 2 m. How much force does he apply? Assume there is no friction.

Physics
2 answers:
Dmitry [639]3 years ago
6 0

The correct answer is A. 250N

Work is a product of force and distance.

That is, work done=force×distance

Therefore substituting for the values in the question:

500J=force×2m

force= 500Nm/2m=250N

another unit for work done is Nm as force as the SI unit of force is newtons and distance in meters

CaHeK987 [17]3 years ago
6 0

Answer:

250

Explanation:

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Why do things become hot or cold?<br> Relate to Thermal Energy and The Law of Conservation of Energy
Mekhanik [1.2K]

<u>Answer:</u>

Things become hot and cold because of the transfer of energy.

<u>Explanation:</u>

The energy possessed by an object or system is called thermal energy and heat is the flow of this energy. While the law of conversation of energy states that energy is not destroyed or created, it just transfers from one object to another.  

When a hot object is placed in normal conditions, it transfers heat to the environment until both are at the same temperature and heat transfers from the environment to the cold objects placed in normal conditions.

4 0
3 years ago
The normal eye, myopic eye and old age
yanalaym [24]

Answer:

1)    f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

        \frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}

        f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

           1 / f = 1/15 + 1 / 1.5

           1 / f = 0.733

            f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

         f’₀ / f = 1.5 / 1.36

         f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

          a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

          tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

          sin θ = y / f

where f is the distance of the lens (eye)

           y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

           y / f = 1.22 λ / D

           y = 1.22 λ f / D

where D is the diameter of the eye

          D = 2R₀

          D = 2 0.1

          D = 0.2 cm

           

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

         

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

      \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

       \frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}

       \frac{1}{f}= 0.7066

        f = 1.415 cm

therefore the diffraction is

        y = 1.22  550 10⁻⁹  1.415  / 0.2

        y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

       d = 2y

       d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

       \frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}

        \frac{1}{f}= 0.733

         f = 1.36 cm

diffraction is

        y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

        y = 4.56 10-6 m

the diffraction diameter is

        d_myope = 2y

         d_myope = 9.16 10-6 m

         \frac{d_{normal}}{d_{myope}} = 9.49 /9.16

        \frac{d_{normal}}{d_{myope}} =  1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0
3 years ago
Lucas left a metal bowl and a wooden bowl outside overnight. The next morning, he picked up the bowls to bring them inside. He n
kipiarov [429]

Answer: C. Metal transfers heat away from the skin by conduction, creating the sensation of coolness.

Explanation: The skin releases heat into the metal bowl since there is a difference in temperature between the two objects. So heat is taken away from the hand abd transfers into the metal bowl by conduction creating a cooler sensation.

8 0
3 years ago
PLEASE HELP ME WILL GIVE BRAINLIEST
weqwewe [10]

Answer:

A.always changing

Explanation:

7 0
3 years ago
Read 2 more answers
What is the expected wavelength (in cm) of 10.5 ghz microwaves in free space?
Sindrei [870]

Wavelength = (speed) / (frequency)

Wavelength = (300 thousand km per second) / (10.5 billion per second)

Wavelength = (300 / 10.5) (thousand km per second) / (billion per second)

Wavelength = (28.57) (million meters / second) / (thousand million / second)

Wavelength = (28.57) (meters / second) / (thousand / second)

Wavelength = (28.57) (meters / thousand)

<em>Wavelength = (28.57) (millimeters) </em>

5 0
3 years ago
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