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bekas [8.4K]
4 years ago
14

The author’s feeling about a subject or topic, which is evidenced in word choice, is called __________.

Physics
2 answers:
True [87]4 years ago
6 0

The author’s feeling about a subject or topic, which is evidenced in word choice, is called tone.

MatroZZZ [7]4 years ago
5 0
The author’s feeling about a subject or topic, which is evidenced in word choice, is called tone. <span>They could be literal meanings which are meanings explicitly given to the readers or the person being told to. In this type of meaning, the person doesn’t need to think the words too much or ponder on the meaning because they could easily understand what the other person or the author implies. Meanwhile connotative meanings are meanings using different figures of speech or symbolism that the person still needs to think it over before knowing the actual meaning of the word, sentence or story.</span>
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Five marbles roll down a ramp. Each marble reaches the bottom of the ramp at a speed of 3 meters/second. Which marble has the hi
Mashutka [201]
The smallest marble has the most kinetic energy therefore it would be the correct answer
8 0
3 years ago
Read 2 more answers
Two identical isolated conducting spheres are picked for an experiment. Only one of them is charged. If the spheres are now brie
aev [14]
Because they are conducting, when you bring them together the charge is split equally among the two spheres (because they have the same radius the amount of charge is also equal). Now they will repel each other because of the net charge on each with the same polarity.
5 0
3 years ago
Does lightning McQueen have car or life insurance?
shutvik [7]
I am taking an educated guess that he has life insurance
7 0
3 years ago
Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid
algol13

Answer:

The intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

|E_{1}|=k\frac{q_{1}}{d^{2}}

Where:

  • k is Coulomb's constant
  • q(1) is 20.00 μC or 20*10⁻⁶ C
  • d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}

|E_{1}|=1.8*10^{7}\: N/C

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

|E_{2}|=k\frac{q_{2}}{d^{2}}

|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}

|E_{2}|=7.2*10^{6}\: N/C

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

I hope it helps you!

5 0
3 years ago
78 grams of a radioactive nuclei X undergoes radioactive decay. The half-life of X is 4.7 minutes. After 16.5 minutes, the remai
kipiarov [429]

<u>Answer:</u> The remaining sample of X is 6.9 grams.

<u>Explanation:</u>

All the radioactive reactions follow first order kinetics.

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=4.7min

Putting values in above equation, we get:

k=\frac{0.693}{4.7min}=0.147min^{-1}

The equation used to calculate time period follows:

N=N_o\times e^{-k\times t}

where,

N_o = initial mass of sample X = 78 g

N = remaining mass of sample X = ? g

t = time = 16.5 min

k = rate constant = 0.147min^{-1}

Putting values in above equation, we get:

N=78\times e^{-(0.147min^{-1}\times 16.5min)}\\\\N=6.9g

Hence, the remaining amount of sample X is 6.9 g

4 0
4 years ago
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