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Dmitriy789 [7]
3 years ago
15

Find the vector parallel to the resultant of the vector A=i +4j-2k and B=3i-5j+k​

Physics
2 answers:
Zinaida [17]3 years ago
8 0

Answer:

2008

Explanation:

2000+3+5======2008

valkas [14]3 years ago
6 0

Answer:

8\hat i-2\hat j-2\hat k

Explanation:

<u>Vectors in 3D</u>

Given a vector

\vec P = P_x\hat i+P_y\hat j+P_z\hat k

A vector \vec Q parallel to \vec P is:

\vec Q = k.\vec P

Where k is any constant different from zero.

We are given the vectors:

\vec A = \hat i+4\hat j-2\hat k

\vec B = 3\hat i-5\hat j+\hat k

It's not specified what the 'resultant' is about, we'll assume it's the result of the sum of both vectors, thus:

\vec A +\vec B = \hat i+4\hat j-2\hat k + 3\hat i-5\hat j+\hat k

Adding each component separately:

\vec A +\vec B = 4\hat i-\hat j-\hat k

To find a vector parallel to the sum, we select k=2:

2(\vec A +\vec B )= 8\hat i-2\hat j-2\hat k

Thus one vector parallel to the resultant of both vectors is:

\mathbf{8\hat i-2\hat j-2\hat k}

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How much heat is released upon converting one mole of steam (18.0
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4.65 × 10⁴ Joules of heat is released upon converting one mole of steam to water.

\texttt{ }

<h3>Further explanation</h3>

Specific Heat Capacity is the amount of energy needed to raise temperature of 1 kg body for 1°C.

\large {\boxed{Q = m \times c \times \Delta t} }

<em>Q = Energy ( Joule )</em>

<em>m = Mass ( kg ) </em>

<em>c = Specific Heat Capacity ( J / kg°C ) </em>

<em>Δt = Change In Temperature ( °C )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial temperature of steam = t = 100.0°C

specific heat capacity of water = c = 4.186 J/gK

mass of steam = m = 18.0 gram

final temperature of water = t' = 25.0°C

specific latent heat of vaporization of water = Lv = 2268 J/g

<u>Asked:</u>

heat released = Q = ?

<u>Solution:</u>

\boxed {\large {\texttt{steam 100}^oC \overset{[Q_1]}{\rightarrow} \texttt{water 100}^oC \overset{[Q_2]}{\rightarrow} \texttt{water 25}^oC}}

Q = Q_1 + Q_2

Q = mL_v + mc\Delta t

Q = 18.0(2268) + 18.0(4.186)(100-25)

Q = 40824 + 5651.1

Q = 46475.1 \texttt{ Joules}

Q \approx 4.65 \times 10^4 \texttt{ Joules}

\texttt{ }

<h3>Learn more</h3>
  • Efficiency of Engine : brainly.com/question/5597682
  • Flow of Heat : brainly.com/question/3010079
  • Difference Between Temperature and Heat : brainly.com/question/3821712

\texttt{ }

<h3>Answer details </h3>

Grade: College

Subject: Physics

Chapter: Thermal Physics

\texttt{ }

Keywords: Heat , Temperature , Block , Aluminium , Ice , Cold , Water

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