1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Step2247 [10]
3 years ago
5

What other forces affect the soccer ball after the student's foot stops touching it?

Physics
2 answers:
hodyreva [135]3 years ago
8 0
Gravity, acceleration, velocity
svet-max [94.6K]3 years ago
4 0
Gravity I hope this helps you I'm not very good at this kind of stuff
You might be interested in
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
A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha
umka21 [38]

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

R_{1} = 330\ohm

R_{2} = 470\ohm

R_{3} = 220\ohm

Since, R_{2} = 470\ohm and R_{3} = 220\ohm are in parallel and this combination is in series with R_{1} = 330\ohm, so,

Equivalent resistance of the circuit is given by:

R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}

R_{eq} = \frac{470\times 220}{470 + 220} + 330

R_{eq} = 149.85 + 330 = 479.85 \ohm

power drain on an ideal battery, P = \frac{V^{2}}{R_{eq}}

                                                      P = \frac{2.9^{2}}{479.85}

                                                      P = 0.017 W

4 0
3 years ago
How can i find the acceleration?(rope and grinder have no weight) *** sorry for my english
Finger [1]
The main formula to be used here is

                       Force = (mass) x (acceleration).

We'll get to work in just a second.  But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them.  So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.

On the <u>right</u> side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration.  I don't know what to do with that, because . . .

At the <em>same time</em>, both of those will be pulled <u>up</u> by the 10 kg on the other side
of the upper pulley.

I think I can handle the 10 kg, and work out the acceleration that IT has.

Let's look at only the forces on the 10 kg:

-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.

-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope. 

So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.

The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.

The acceleration of 10 kg, with 49 newtons of force on it, is

     Acceleration = (force) / (mass) = 49/10 = <em>4.9 meters per second²</em>
7 0
3 years ago
What leads astronomers to believe that some large moons associated with the giant planets have compositions that are roughly hal
Sonja [21]
<span>Astronomers are able to determine facts about the composition of these moons by examining the nature of light that is reflected from their surfacy using a method called spectroscopy. This process works because different materials tend to reflect light at different wavelengths So, by observing at which wavelengths a planetary body reflects light, astronomers are able to estimate its composition.</span>
8 0
3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
2 years ago
Other questions:
  • Newton's first law states that an object traveling with a constant velocity will remain traveling at a constant velocity unless
    13·1 answer
  • Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t
    5·1 answer
  • A particle's position along the x-axis is described by. x(t)= At+Bt^2where t is In seconds: x is in meters: and the constants A
    11·1 answer
  • The velocity of in water is 1500 m/s.
    5·1 answer
  • If a baseball travels a distance of 4 meters in 5 seconds, what is its average speed
    15·1 answer
  • greater than: The electric potential energy of a proton at point A is _____ the electric potential energy of an proton at point
    8·1 answer
  • Suppose that you are holding a cup of coffee in your hand identify all forces on the cup and reaction to each force
    6·2 answers
  • Describe an object that emits radiation in Astronomy
    9·1 answer
  • An object with a height of 1.0 cm is placed 1.4 cm from a convex lens that has a
    10·1 answer
  • Which image shows a nonrenewable resource?<br><br> sun<br> fire<br> can<br> plant
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!