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Zielflug [23.3K]
2 years ago
14

While traveling along a highway a driver slows from 31 m/s to 15 m/s in 8 seconds. What is the automobile’s acceleration? (Remem

ber that a negative value indicates a slowing down or deceleration.)​
Physics
1 answer:
MaRussiya [10]2 years ago
8 0

Answer:

The automobile's acceleration in that time interval is -2 m/s^2

Explanation:

The acceleration is defined as the rate of change of the velocity.

The average acceleration in a given lapse of time is calculated as:

A = (final velocity - initial velocity)/time.

In this case, we have:

initial velocity = 31 m/s

final velocity = 15 m/s

time = 8 seconds.

Then the average acceleration is:

A = (15m/s - 31m/s)/8s = -2 m/s^2

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A trombone has a variable length. When a musician blows into the mouthpiece and causes air in the tube of the horn to vibrate, t
Lilit [14]

Answer:

The frequency increases.

Explanation:

When the Musician draws the slide in the length of the horn gets shorter, which causes a decrease in the wavelength. A decrease in the wave length results in an increase in frequency.

Note:

The diameter of the horn has an effect on frequency, so a wider horn is effectively a long horn - open end correction ( distance between the the antinode and the open end of a pipe).

Frequency also depends on how hard the musician blows the trombone. The musician can change the frequency with the lip pressure being applied.

6 0
3 years ago
A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t
kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta (1)

where

F is the magnitude of the force

d is the distance covered by the object

\theta is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

\theta=0

And the distance covered is equal to the circumference of the circle, which is:

d=2\pi r=2\pi (2.5 m)=15.7 m

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

W=(3.0)(15.7)(cos 0)=47.1 J

Learn more about work:

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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
When Mendeleev organized elements in his periodic table in order of increasing mass, similar elements with similar properties we
Temka [501]
Similar elements with similar properties were in the same groups and periods for instance lithium(Li) and sodium(Na) are alkaline metals and so belong to the same group (that is group 1).Also Hydrogen(H) and Helium(He) both have only one shell or energy level and so belong to the same period.
8 0
3 years ago
Give one example of a thermodynamic cycle that does not account for the carnot efficiency.
Arturiano [62]

Thermo-Electrochemical converter (UTEC) is a thermodynamic cycle that does not account for the Carnot Efficiency.

The Carnot cycle is a hypothetical cycle that takes no account of entropy generation. It is assumed that the heat source and heat sink have perfect heat transfer. The working fluid also remains in the same phase, as opposed to the Rankine cycle, in which the fluid changes phase. A practical thermodynamic cycle, such as the Rankine cycle, would achieve at most 50% of the Carnot cycle efficiency under similar heat source and heat sink temperatures.

<h3>What is Thermo-Electrochemical converter?</h3>

In a two-cell structure, a thermo-electrochemical converter converts potential energy difference during hydrogen oxidation and reduction to heat energy.

It employs the Ericsson cycle, which is less efficient than the Carnot cycle. In a closed system, it converts heat to electrical energy. There are no external input or output devices.

This means there will be no mechanical work to be done, as well as no exhaust. As a result, Carnot efficiency is not taken into account in this cycle. Carnot efficiency is accounted for by other options such as turbine and engine.

Learn more about Thermo-Electrochemical converter here:

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4 0
2 years ago
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