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3 years ago

Why are velocity and acceleration called vector quantities?

Physics

2 answers:

Airida [17]3 years ago

4 0

Answer:because they both have magnitude and direction. Which can be identified from their SI Units.

Explanation:

because they both have magnitude and direction. Which can be identified from their SI Units.

Stella [2.4K]3 years ago

4 0

Answer:

Acceleration is a vector quantity because it has both magnitude and direction. When an object has a positive acceleration, the acceleration occurs in the same direction as the movement of the object.a change in the direction of motion results in an acceleration even if the moving object neither sped up nor slowed down. That's because acceleration depends on the change in velocity and velocity is a vector quantity — one with both magnitude and direction.Velocity as a Vector Quantity Because the person always returns to the original position, the motion would never result in a change in position. Since velocity is defined as the rate at which the position changes, this motion results in zero velocity.

Explanation:

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Lidia makes a graphic organizer to compare temporary magnets with permanent magnets.

Nitella [24]

Labels that belong in regions marked X and Y are;

A) X: Maintains magnetic properties in the presence of another magnet

Y: Is made from magnetically hard ferromagnetic material

7 0

3 years ago

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Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

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4 years ago

What happens to the orbit time as the distance increases.

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The orbital period increases if the orbital distance is increased.

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3 years ago

Which only list metalloids

HACTEHA [7]

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Explanation:I know this cuz I had this question on my unit test and I had to look it up online....Hope this helps

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3 years ago

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Tcecarenko [31]

So acceleration = (final velocity - initial velocity)/time

So (fv-iv)/t=a

(45-110)/4.5

Gives you (-130/9)km/h^2

You may have to convert the SI units so just follow my steps and change what must be changed

4 0

3 years ago

Read 2 more answers