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Ugo [173]
3 years ago
11

Which element is not part of the rhetorical triangle designed by Aristotle? speaker credibility audience emotion content logic l

ength of address
Physics
2 answers:
sdas [7]3 years ago
7 0

The correct answer is:

D. Length of address.

The length of address is not part of the rhetorical triangle designed by Aristotle.

|Huntrw6|

ser-zykov [4K]3 years ago
3 0
The correct answer is this one: "length of address." The element that is not part of Aristotle's rhetorical triangle is the length of the address. The rhetorical triangle of Aristotle are the logos, pathos, and ethos. Logos is the logic, Pathos is the audience emotion, Ethos is the speaker's credibility.
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An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

4 0
2 years ago
If the unit for force is f, the unit for velocity is v and the unit for time t, then the unit for momentum is
ASHA 777 [7]
The answer is m I believe
5 0
3 years ago
A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed
AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

\omega=\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s

And the frequency of the motion instead is given by:

f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

F=kA

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N

k = 160 N/m is the spring constant

Solving for A, we find

A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m

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Answer:

The answer is 169253.55 joules

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