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sertanlavr [38]

3 years ago

6

What is the purpose of the author’s description of Montresor’s encounter with Fortunato above? The author is giving the reader a

background of the friendship between these two characters. This description is included as a part of the setting of the story The author is contrasting the innocence of Fortunato with the maliciousness of Montresor. The author is foreshadowing the misfortune that will happen to Fortunato.

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

If my understanding is right I would say the answer is. D cause foreshadowing means what is to come in the future. He talks about future events of punishment to come. Hope this helps.

evablogger [386]3 years ago

4 0

Well if the encounter is good then that would be great to write about!!

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The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton

Brrunno [24]

Answer:

A) F_g = 4.05 10⁻⁴⁷ N, B) F_e = 9.2 10⁻⁸N, C) $\frac{F_e}{F_g}$ = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

F = $G \frac{m_1m_2}{r^2}$

let's calculate

F = $6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }$

F_g = 4.05 10⁻⁴⁷ N

B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

F = $k \ \frac{q_1q_2}{r^2}$

let's calculate

F = $9 \ 10^9 \ \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}$

F_e = 9.2 10⁻⁸N

C) It is asked to find the relationship between these forces

$\frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }$

= 2.3 10³⁹

therefore the electric force is much greater than the gravitational force

4 0

3 years ago

Keisha finds instructions for a demonstration on gas laws. 1. Place a small marshmallow in a large plastic syringe. 2. Cap the s

lana [24]

The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
</span><span>
Keisha follows the instructions for a demonstration on gas laws.
1. Place a small marshmallow in a large plastic syringe.
2. Cap the syringe tightly.
3. Pull the plunger back to double the volume of gas in the syringe.

Now, this activity is being done at the same temperature, because there is no mention of the temperature change. Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.

7 0

3 years ago

Read 2 more answers

Dalton proposed the first atomic model. Which of these statements accurately reflected his thinking at that time?

Sunny_sXe [5.5K]

B. Atoms are solid balls

3 0

3 years ago

Read 2 more answers

You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame

Novay_Z [31]

Answer:

The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

$\phi=NAB$

Put the value into the formula

$\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5$

$\phi=0.003180$

We need to calculate the emf

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{0.003180}{0.14}$

$\epsilon =-0.000227\ V$

$\epsilon=-2.27\times10^{-4}\ V$

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

8 0

3 years ago

A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between

Monica [59]

Answer:

a) $t=5.5\mu s$

b) $\Delta t'=12.5\mu s$

Explanation:

a) Let's use the constant velocity equation:

$v=\frac{\Delta x}{\Delta t}$

v is the speed of the muon. 0.9*c
c is the speed of light 3*10⁸ m/s

$t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}$

$t=5.5\mu s$

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

$\Delta t'=\gamma\Delta t$

$\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

v is the speed of muon (0.9c)

Therefore the time in the rest frame will be:

$\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t$

$\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t$

$\Delta t'=\frac{1}{0.44}\Delta t$

No we use the value of Δt calculated in a)

$\Delta t'=2.27*5.5=12.5\mu s$

I hope it helps you!

8 0

3 years ago