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defon
3 months ago
9

Draw the vector C⃗ =1.5A⃗ −3B⃗ . The length and orientation of the vector will be graded. The location of the vector is not impo

rtant.

Physics
2 answers:
Yuliya22 [3.3K]3 months ago
7 0
I created the illustration found in the accompanying file.

There are two images included.

The upper one illustrates the impacts of:

- scaling vector A by a factor of 1.5, depicted in red with a dashed line.

- scaling vector B by -3, shown in purple with a dashed line.

The lower image displays the resultant vector: C = 1.5A - 3B.

The approach involves relocating the tail of vector -3B to the tip of vector 1.5A while maintaining the angles.

Next, an arrow is drawn from the tail of 1.5A to the position of -3B after this shift.

The arrow representing the result is vector C, marked with a black dashed line.
 
ValentinkaMS [3.4K]3 months ago
5 0

Additional details

A vector is a dimensional quantity characterized by both magnitude and direction.

Vectors can be represented by directed segments.

\large{\boxed{\bold{\overrightarrow{A}}}}

Here, the magnitude of a vector is expressed as | a |.

Vectors can be outlined as coordinate pairs indicating their locations within the Cartesian coordinate system:  a (a₁, a₂).

with length \large{\boxed{\bold{|a|\:=\:\sqrt{a_1^2+a_2^2}}}}

Reversing the direction of vector a yields the vector -a, which retains the same magnitude yet points in the opposite direction.

Vector operations encompass adding and subtracting. The addition of vector a and vector b can be accomplished using a triangular method, aligning the start point of vector b with the end of vector a.

The result is achieved by drawing a segment from the beginning of vector a to the end of vector b, which creates a new vector c.

Thus, we have a + b = c.

When you add vector a to the negative of vector b (-b), the equation becomes a + (-b) = a - b.

Multiplying a vector by a scalar (represented as k) transforms it to k | a |.

If k > 0, the new vector points in the same direction as vector a, whereas if k < 0, it goes in the opposite direction.

The problem provides a partial diagram (see attached), and the total vector to find is C = 1.5A - 3B.

Scaling vector A by 1.5 results in a length that is 1.5 times that of vector A and maintains the same direction. Likewise, vector B, affected by -3, has a length that is three times that of vector B but points in the opposite direction. Hence, we derive vector C (refer to attached image) by drawing a line from the starting point of A to the endpoint of B.

Explore further

velocity position

understanding vector components

coherence in vector operations

Keywords: vector, addition, subtraction

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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
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Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

\texttt{ }

Additional explanation

The Ideal Gas Law that should be remembered is:

\large {\boxed {PV = nRT} }

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

\texttt{ }

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

Next, we will determine the work performed on the gas:

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

Step B:

By utilizing the methodology mentioned earlier:

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

Next, we will ascertain the work completed on the gas:

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

Ultimately, we can calculate the total work done and heat supplied as follows:

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

Learn more

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\texttt{ }

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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