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Alexandra [31]
2 years ago
8

Help!!!! please!!! i’m not sure what the answer is for both questions please

Physics
1 answer:
n200080 [17]2 years ago
7 0

Answer:

a.) 600N

b.) 2 m/s^2

Explanation:

a.)

In order to solve for the needed force, you must use Newton's 2nd law which states that a force is equal to mass multiplied by acceleration.

Basically: Force = mass * acceleration ; F = ma

Since both mass and acceleration are already given, just plug them into the formula and solve for force:

Mass = 200kg

Acceleration = 3m/s^2

F = m * a = 200 * 3 = 600kg * m/s^2 = 600N

b.)

In order to find the acceleration, like the hint already says, you need to find the net force first and then use Newton's 2nd law.

To calculate net force:

Since both forces are parallel (affect the same axis) and in opposite directions, you can subtract the two in order to find resultant magnitude. The resultant direction is the direction of the force with the large magnitude.

Net Force: 8N - 4N = 4N

Net Force Direction: since 8N > 4N, the direction is in the same direciton as the 8N force = Left

Finally, you find acceleration using newton's 2nd law, F = ma.

Since you already know mass and net force, plug in to solve for acceleration:

4N = 2kg * a

a = 4 N / 2kg = 2m/s^2

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Let <em>A</em> be the amount of money that Aliya deposited in the savings account. Since <em>A</em> 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 <em>2A.</em>

<em />

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

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 <em>2A</em>, 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 <em>A</em> 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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Explanation:

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The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1
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The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

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