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
Olenka [21]
3 years ago
13

A different group of students decided to do this experiment by

Physics
1 answer:
GaryK [48]3 years ago
8 0

Answer:

Using an object of less mass can reduce the energy lost due to friction.

Explanation:

let the mass of the object be M.

acceleration due to gravity be g, coefficient of kinetic friction between object and slider be \alpha  , angle between horizntal and slider be ∅ and distance slided by the object be d.

work done by friction of the ramp will be the energy lost by ramp due to friction.

work done = force×displacement

force = \alpha×M×g×cos∅

displacement= d

therefore work done =  \alpha×M×g×cos∅×d

therefore on decreasing the mass M we can see that work done by friction decreases so the energy lost due to friction also decreases.

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
50200 J of heat are removed from
Dmitry_Shevchenko [17]

Correct Answer:

3.1375

Explanation:

Use equation Q=mcΔT to find m

Plug in all variables -50200=x\cdot 2000\cdot -8

Answer: 3.1375

4 0
3 years ago
An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to
AURORKA [14]

Answer:

Explanation:

Expression for relative velocity

= \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }

= (.54 + .82 )c/ 1+ \frac{.54 \times.82}{1}

= 1.36 c / 1.4428

= .94 c

β = .94

6 0
3 years ago
A block is attached to a horizontal spring. On top of this block rests another block. The two-block system slides back and forth
Minchanka [31]

Answer:

Angular frequency will increase

No change in the amplitude

Explanation:

At extreme end of the SHM the energy of the SHM is given by

E = \frac{1}{2} (m_1 + m_2)\omega^2 A^2

here we know that

\omega^2 = \frac{k}{m_1 + m_2}

now at the extreme end when one of the mass is removed from it

then in that case the angular frequency will change

\omega'^2 = \frac{k}{m_1}

So angular frequency will increase

but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block

so here no change in the amplitude

6 0
3 years ago
For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of
natali 33 [55]

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

d = k t^2

where k is the constant of proportionality in furlongs / s^2

 

<span>Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:</span>

2 = k (2)^2

k = 2 / 4

k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

 

Now solving for d when t = 4:

d = 0.5 (4)^2

d = 0.5 * 16

<span>d = 8 furlongs</span>

<span>
</span>

<span>It traveled 8 furlongs for the first 4.0 seconds.</span>

8 0
3 years ago
Other questions:
  • Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid ra
    10·1 answer
  • The following graph shows the rates of decay for four parent isotopes. Based on the graph, which parent isotope most likely has
    11·2 answers
  • A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans
    14·2 answers
  • How could the position data from a motion sensor be used to create a velocity--time graph?
    8·1 answer
  • A man begins pushing a lwan mower across a lawn from a standing start. Which is true? A. The man is decreasing the kinetic energ
    9·1 answer
  • Hydraulic systems utilize Pascal's principle by transmitting pressure from one cylinder (called the primary) to another (called
    7·1 answer
  • The differences between models and scientific laws and theories.
    11·1 answer
  • Please help, only answer if your 1000% correct im in summer school and need to pass this class
    15·2 answers
  • Helpp
    13·1 answer
  • How much does a 60 kg table weigh
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!