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3 years ago

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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2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

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So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

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$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

2 years ago

What is the gpe of a 200 kg hot air ballon 21,000 m above the ground?

stiv31 [10]

gravitational Potential energy is given by

$GPE = mgh$

here we have

m = 200 kg

h = 21000 m

now we will have

$GPE = 200(9.8)(21000)$

$GPE = 4.116 \times 10^7 J$

so GPE of balloon will be 41160000 J above the given height from ground

8 0

3 years ago

suppose a ball is thrown vertically upward. Eight seconds later it returns to its point of release. What is the initial velocity

valentinak56 [21]

The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.

At its highest point, its velocity changed from upward to downward.
At that instant, its velocity was zero.

The acceleration of gravity is 9.8 m/s². That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second.

-- If the object is falling downward, it moves 9.8 m/s faster every second.

-- If the object is tossed upward, it moves 9.8 m/s slower every second.

The ball took 4 seconds to lose all of its upward speed. So it must have
been thrown upward at (4 x 9.8 m/s) = 39.2 m/s .

(That's about 87.7 mph straight up. Somebody had an amazing pitching arm.)

6 0

3 years ago

The combination of all of the forces acting on an object is called the

Veronika [31]

The answer is Net force

6 0

3 years ago

A bullet of mass 0.01 kg is fired in to a sand bas of mass 0.49 kg from a tree. The Sand bag with the bullet embedded in to it s

makvit [3.9K]

Since the bag was at rest, its initial momentum is zero. The velocity of the ball before collision is 500 ms-1.

<h3>Linear momentum</h3>

The term momentum in physics refers the product of mass and velocity. If we know mass of the object and its velocity, then we calculate the momentum.

Momentum before collision for the bullet = 0.01 kg × v

Momentum before collision for the bag = 0

Momentum after collision for the bag and bullet = (0.01 kg + 0.49 kg) 10 = 5 Kgms-1

The velocity of the bullet before collision = 0.01 kg × v + 0 = 5 Kgms-1

v = 5 Kgms-1/0.01 kg

v = 500 ms-1

Learn more about momentum: brainly.com/question/904448

8 0

2 years ago

Read 2 more answers

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

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