Vectors are arrows that tell us two things about force...WEIGHT and DIRECTION

14. Determine the kinetic energy of a 1000-kg roller coaster car that is moving with a speed of

ElenaW [278]

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 1000 \times {40}^{2} \\ = 500 \times 400$

We have the final answer as

Hope this helps you

6 0

3 years ago

Read 2 more answers

Vector A has magnitude of 8units and makes an angle of 45° with the positive x-axis. Vector B also has the same magnitude of 8un

iragen [17]

Answer:

Explanation:

Because vectors have direction and x and y components you can't just add them and say that their length is 16 because A is 8 units and so is B. What you're actually finding is the magnitude and direction of the vector that results from this addition. The magnitude is the length of the resultant vector, which comes from the x and y components of A and B, and the direction is the angle between the resultant vector and the positive x axis. To add the vectors, then, we need to find the x and y components of each. We'll do the x components of A and B first so we can add them to get the x component of C. Since x values are directly related to cos, the formula to find the x components of vectors is

$V_x=Vcos\theta$ which is the magnitude of the vector (its length) and the angle. Finding the x components of A:

$A_x=8.0cos45$ so

$A_x=5.7$ and for B:

$B_x=8.0cos180$ since the negative x axis is the 180 degree axis and

$B_x=-8.0$ If we add them, we get

$C_x=-2.3$

Now onto the y components. The formula for that is almost the same as the x components except use sin instead of cos:

$A_y=8.0sin45$ so

$A_y=5.7$ and

$B_y=8.0sin180$ so

$B_y=0$ If we add them, we get

$C_y=5.7$

Now for the final magnitude:

$C_{mag}=\sqrt{(-2.3)^2+(5.7)^2}$ and

$C_{mag}=6.1 units$ and now onto the direction.

The x component of C is positive and the y component is negative, which means that the direction has us at an angle is quadrant 2; we add 180 to whatever the angle is. Finding the angle:

$tan^{-1}(\frac{C_y}{C_x})=(\frac{5.7}{-2.3})$ = -68 + 180 = 112 degrees

The resultant vector of A + B has a magnitude of 6.1 and a direction of 112°

Do the same thing for subtraction, except if you're subtracting B from A, the direction that B is pointing has to go the opposite way. That means that A doesn't change anything at all, but B is now pointing towards 0.

$A_x=5.7$ (doesn't change from above)

$B_x=8.0cos0$ and

$B_x=8.0$ so

$C_x=13.7$ and

$A_y=5.7$ (also doesn't change from above)

$B_y=8.0sin0$ so

$B_y=0$ and

$C_y=5.7$ and for the magnitude:

$C_{mag}=\sqrt{(13.7)^2+(5.7)^2$ so

$C_{mag}=15units$ and for the direction:

$tan^{-1}(\frac{5.7}{13.7})=23$ and since both x and y components of C are in Q1, we add nothing.

And you're done!!!

3 0

3 years ago

A person with myopia (near-sightedness) has a far point of 45.0cm while their near point is 15.0cm. Upon wearing glasses, they h

Crank

Answer:

p = 22.5 cm

Explanation:

For this exercise we must use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image respectively.

Let's start with the far point, the object is very far away (p = ∞) and the image must be formed at the far point of view of the person q = 45.0 cm

since the image is on the same side as the object according to the sign convention the distance is negative

$\frac{1}{f} = \frac{1}{\infty } + \frac{1}{-45}$

f = -45.0 cm

now let's use the near point (q = 15.0 cm) at what distance the object should be

$\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$

$\frac{1}{p} = \frac{1}{-45} - \frac{1}{-15}$ 1 / p = 1 / -45 - 1 / -15

$\frac{1}{p} = - \frac{1}{45} + \frac{1}{15}$ 1 / p = -1/45 + 1/15

$\frac{1}{p}$ = 0.0444

p = 22.5 cm

this is the closest distance you can see an object clearly

4 0

3 years ago

Two identical massless springs of constant k = 200N/m are fixed at opposite ends of a level track, as shown. A 5 kg block is pre

Rina8888 [55]

Energy in spring:
<span>E = ½k · x²; where x = 0.15m </span>

<span>From coefficient of friction, </span>
<span>F = µ · N = µ · mg; where m = 5kg and g = 9.807m/s² </span>

<span>Work done each pass = </span>
<span>W = ∫Fdx = Force · distance </span>

<span>Just divide E by W to see how many passes, use this to figure out where it stops. </span>

5 0

3 years ago

Consider the mechanism. Step 1: 2A↽−−⇀B+C equilibrium Step 2: B+D⟶E slow Overall: 2A+D⟶C+E Determine the rate law for the overal

Cerrena [4.2K]

Answer : The rate law for the overall reaction is, $R=\frac{K[A]^2[D]}{[C]}$