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3 years ago

What is imagery?

2 answers:

8 0

Imagery is almost like a film being played in the readers mind when reading literature. So on that note, I think the last one, 'words that trigger the imagination to remember by appealing to the five senses', is the best choice.

I hope this helps!

Rudik [331]3 years ago

5 0

Correct answer choice is :

D) Words that trigger the imagination to remember by appealing to the five senses

Explanation:

Imagery, in a literary text, is an author's use of clear and expressive voice to add intensity to their work. It advances to human senses to increase the reader's understanding of the work. Powerful forms of imagery involve all of the senses. Basically, there are five types of imagery, each corresponding to one of our senses, visual, auditory, kinesthetic olfactory, and gustatory. The key to good imagery is involving all five senses.

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Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released qu

Kruka [31]

Answer: a) 1766 sec. b) 55.5 MJ c) 13.9 MW d) -12,944 Nm

Explanation:

a) The torque and the angular acceleration, are related by the following expression, which resembles very much to the Newton's 2nd Law for point masses:

ζ = I . γ, where ζ=external torque, I = rotational inertia and γ = angular acceleration.

We also know that a flywheel is a solid disk, so the rotational inertia for this type of body is equal to MR² / 2.

By definition, angular acceleration is the rate of change of angular velocity with time, so we can write the following:

γ = ωf -ω₀ / t

Assuming that the flywhel starts from rest, we know that ω₀ = 0, and ωf = 12,000 rpm.

As all the units are given in SI units, it is advisable to convert the rpm to rad/sec, as follows:

12,000 rpm = 12,000 rev. (2π/rev) . (1min/60 sec) = 400 π rad/sec

Returning to the original equation, we have:

ζ = MR² / 2 . (ωf/ t)

Replacing by the values, and solving for t, we have:

t = 250 Kg. (0.75)² m² . 400 π / 2. 50 Nm = 1,766 sec.

b) Due to the flywheel is just rotating, all the stored energy is rotational kinetic energy, which can be written as follows:

K = 1/2 I ωf² = 1/2 (MR²/2) ωf² = 1/4. 250 Kg. (0.75)² m². (400π)²

K= 55.5 MJ

c) Power is defined as energy delivered in a given time.

The energy delivered, is just the half of the originally stored value, i.e. , 55.5 MJ /2, equal to 27.75 MJ.

Dividing this value by 2.0 sec, we have the average power delivered to the machine, that we found to be equal to 27.75 MJ / 2s = 13. 9 MW

d) Using the same relationship than in a), we can write the following:

ζ = I. γ

I remains the same (as the flywheel is the same), so the only unknown is the angular acceleration.

Angular acceleration, by definition, is as follows:

γ = ωf - ω₀ / t

We know the value of ω₀, as it is the top speed value that we have already got,i.e., 400 π rad/sec.

We don't know the value for ωf, but we know the value of the rotational kinetic energy after 2.0 secs, which is equal to the half of the one we obtained in step b).

So, we can write the following:

Kf = 1/2 I ωf² = 1/2 (1/2 I ω₀²) ⇒ 1/ 2 ωf² = 1/4 ω₀² ⇒ωf = ω₀/√2

Replacing in the expression for angular acceleration:

γ = (ω₀/√2 - ω₀) / t = -0.29. 400. π/ 2 rad/sec²= -184.1 rad/sec²

Finally, we can get the torque as follows:

ζ = (250 kg. (0.75)² m² /2) . 184.1 rad/sec² = -12,944 Nm

6 0

4 years ago

When do you produce more pressure on the ground...standing or laying down?

ycow [4]

When you are Standing you produce more pressure on the ground

7 0

3 years ago

If we decrease the time it takes for a car to travel over the same distance, this will

Ilya [14]

C) increase the speed

7 0

3 years ago

Read 2 more answers

A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the

GenaCL600 [577]

a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

$E=\phi + K$ (1)

where

$E=\frac{hc}{\lambda}$ is the energy of the incident photon, with

h = Planck constant

c = speed of light

$\lambda$ = wavelength

$\phi$ = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

$eV=K$

So we can rewrite (1) as

$E=\phi + eV$

where we have:

$\lambda=200 nm = 2\cdot 10^{-7} m$

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7}m}=9.95\cdot 10^{-19} J$

Converting into electronvolts,

$E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV$

And now we can solve eq.(1) to find the work function of the metal:

$\phi = E-eV=6.22 eV-1.93 eV=4.29 eV$

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so $\phi$ is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

6 0

4 years ago

A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per

FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0

3 years ago