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3 years ago

How can vectors be used in real life?

1 answer:

4 0

Using vectors can be troublesome for simple physics problems. however, when you move onto solving more difficult problems that have several variables, and the space you are dealing with is in three dimensions, using vectors is a necessity because it's not always possible to visualize the problem. but using vectors will give you a definitive answer.

few applications of vectors:
-predict the trajectory of airplanes and make sure that the airplane will not collide with some other airplane
-all simulations use vectors
-multivariable Calculus is pretty much based on vectors. without multivariable Calculus, it would have not been possible to compute many formulas.
-modeling the path a rocket will take, and supplying it the right amount of fuel so that it gets to the moon, or some other destination.

You might be interested in

A sled of mass 8 kg slides along the ice. It has an initial speed of 4 m/s but

marysya [2.9K]

As our story begins, the sled ... whose mass is 8 kg ... is sliding along the ice at a speed of 4 m/s.

The sled's kinetic energy is (1/2 m v²) = (4 kg · 16 m²/s²) = 64 J .

After what seems like only the blink of an eye, the sled is no longer sliding. It is stationary. Motionless. At Rest. Just sitting there !

Its speed has been reduced to zero and ... because kinetic energy is the energy of motion ... the sled's kinetic energy is now also zero. Sixty-four Joules of energy have disappeared !

How can this be ? ! ? We know that energy is conserved. It can never just appear out of nothing, and it can never just disappear into nothing. If energy suddenly appears, it had to come from somewhere, and if energy suddenly disappears, it had to go somewhere. So where did our 64 Joules of kinetic energy go ?

It went into the ice, THAT's where ! We can say that the sled did 64J of work, and melted a thin slick layer of water on the surface of the ice. OR we can say that friction did NEGATIVE 64J of work on the sled, to cancel the 64J that it had originally, sap its kinetic energy, and bring it to rest.

I think <em>choice-B</em> was supposed to say "<em>B. -64J</em>", but somebody typed it sloppily and neglected to proofread it before posting.

6 0

3 years ago

I need help with one question on my homework. This is on the Specific Heat Capacity required practical.

Leviafan [203]

Answer:

will mostly accord at the top of the boiling water my kind sir

Explanation:

Evaporation takes place only at the surface of a liquid, whereas boiling may occur throughout the liquid. In boiling, the change of state takes place at any point in the liquid where bubbles form. The bubbles then rise and break at the surface of the liquid.

5 0

3 years ago

When designing a user interface, the most important information should be placed in the ______ of the screen.

musickatia [10]

Answer:

upper-left corner

Explanation:

Most vital information are positioned in a place where users can view them clearly and without obstruction.

3 0

3 years ago

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

I am Lyosha [343]

Answer:

$2.09\ \text{m/s}$

$22298.4\ \text{J}$

Explanation:

m = Mass of each the cars = $1.6\times 10^4\ \text{kg}$

$u_1$ = Initial velocity of first car = 3.46 m/s

$u_2$ = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

$mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}$

Speed of the three coupled cars after the collision is $2.09\ \text{m/s}$ .

As energy in the system is conserved we have

$K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}$

The kinetic energy lost during the collision is $22298.4\ \text{J}$ .

6 0

3 years ago

he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :

$\sf{\omega}$ = angular frequency = ... rad/s

Step by step :

$\sf{\omega = \frac{\theta}{t}}$

$\sf{\omega = \frac{2,000 \pi}{54.9}}$

$\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}$

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0

2 years ago