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3 years ago

PLEASE HELP ME!!!

Physics

1 answer:

s344n2d4d5 [400]3 years ago

3 0

Answer:

I think is B. sound, nuclear, electrical

but if I wrong then I am so sorry

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Aliya deposited half as much money in a savings account earning 1.9% simple interest as she invested in a money market account t

yanalaym [24]

Let A be the amount of money that Aliya deposited in the savings account. Since A is half as much as money as she invested in a money market account, then the amount that she invested in the market account is 2A.

Express the interest that Aliya earned in terms of A. Set it equal to the amount of $297.60 and then solve for A.

Since the savings account gives 1.9% simple interest, the total amount of interest that she will earn from the savings account is 1.9% of A, which is equal to:

$\frac{1.9}{100}\times A$

Since the money market account gives 3.7% simple interest, the total amount of interest that she will earn from the money market account, is 3.7% of 2A, which is equal to:

$\frac{3.7}{100}\times2A$

Add both interests in terms of A and simplify the expression:

$\begin{gathered} \frac{1.9}{100}\times A+\frac{3.7}{100}\times2A \\ =\frac{1.9}{100}\times A+\frac{7.4}{100}\times A \\ =(\frac{1.9}{100}+\frac{7.4}{100})\times A \\ =\frac{1.9+7.4}{100}\times A \\ =\frac{9.3}{100}\times A \end{gathered}$

The expression (9.3/100)*A represents the total interest after one year. Then:

$\begin{gathered} \frac{9.3}{100}\times A=297.60 \\ \Rightarrow A=\frac{100}{9.3}\times297.60 \\ \Rightarrow A=\frac{100\times297.60}{9.3} \\ \Rightarrow A=\frac{29760}{9.3} \\ \Rightarrow A=3200 \end{gathered}$

Use the value of A to find the amount that was invested in the money market account:

$2A=2\times3200=6400$

Therefore, Aliya deposited 3200 in a savings account and 6400 in a money market account.

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1 year ago

You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

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Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, $\omega = 25.13 rad/s$

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = $\frac{1}{T}$

f = $\frac{1}{0.25}$

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

$\omega = 2\pi \times f$

$\omega = 2\pi \times 4$

$\omega = 25.13 rad/s$

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