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3 years ago

For Jane to see an image, light must enter her eyes. What specifically is entering her eyes when she sees an image?

Physics

2 answers:

Hitman42 [59]3 years ago

6 0

the answer is u÷7÷3=76

Sergio [31]3 years ago

3 0

Energy
because once the light hits her eyes energy flows through her body so the answer is A energy

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I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

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5 0

3 years ago

Explain three ways visual aids help you study

Katen [24]

Visual aids are tools that help to make an issue or lesson clearer or easier to understand and know (pictures, models, charts, maps, videos, slides, real objects etc.). ... Visual aids are those devices which are used in classrooms to encourage students learning process and make it easier and interesting.

7 0

3 years ago

The bending of light as it passes into a transparent material of different optical intensity is known as A. conversion. B. aberr

stiks02 [169]

Answer:

D. refraction

Explanation:

Refraction: Refraction is change in direction of light rays. Refraction occurs whenever light rays travels from a transparent medium to another transparent medium of different density. The abrupt change in direction at the surface of the surface of the two media is referred to as refraction.

Refraction occurs when light travels from air to glass or from air to liquid.

Laws Of Refraction:

(i) The incident ray, the refracted ray and the normal, all at the point of incident lies in the same plane.

(ii) The ratio of the sine of the angle of incident to the sine of the angle of refraction is a constant for a given pair of media.

Thus the right option is D. refraction

6 0

3 years ago

20 Points available for physics help

Sonja [21]

The second one is correct not sure about the first one sorry

8 0

3 years ago

Read 2 more answers

The volume of gas in a flexible container at a depth of 10 m/33 ft will expand to ______ its original volume if taken to the sur

natulia [17]

The volume of gas in a flexible container at a depth of 10 m/33 ft will expand to double its original volume if taken to the surface.

What is volume:

The amount of space that a substance or object occupies.

Volume is denoted with letter V

What is a gas molecule?

Generally, a gas molecule is simply defined as a combination of numbers of atoms that are connected one to another.

here, 10 m/33 feet of sea water is taken to exerts a gauge pressure that is the same pressure as the atmosphere.

This means that we add one atmosphere/bar pressure for every 10 m/33 feet you descend.

The effect of this is that,

at 10 m/33 ft absolute pressure is two ata/bar

Adding another 10 m/33 ft to this depth to make it 20 m/66 ft, puts you under under three ata/bar of pressure and so on.

Descending to a depth it means that you have double the ata/bar pressure acting on you. If you decide to resurface, the pressure on the flexible container will reduce by a factor of two.

Since volume is inversely proportional to the pressure,

If the pressure reduces by a factor of 2, then the volume will increase by a factor of two.

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3 0

2 years ago