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melamori03 [73]
3 years ago
12

The laboratory exercise for this chapter addresses kinematic motion. You have experienced these motions in everyday life. Instea

d of a discussion board requiring posts in a forum, this assignment has been modified to accept your response to the following questions in this assignment. Be sure to clearly address each of the points below and show all of your work • What is the difference between a vector and a scalar quantity? • List two examples each of vector and scalar quantities • Write the due date of this assignment: Month and Day (For example, July 15 would be Month - 7. Day = 15) • For a building having a height equal to the quantity you have recorded for Day in meters in our example 15 meters), compute the time required for the ball to fall to the ground while experiencing acceleration due to gravity (g=9.8m/s) • How fast was the ball traveling when it hit the ground? Submit the kinetics Assignment by 11:59 p.m. (ET) on Monday,
Physics
1 answer:
ki77a [65]3 years ago
5 0

Answer:

A) vectors: veloicty, force

scalar:  speed, work

B)  t = 1.75 s,  C)   v = - 17 2 m / s

Explanation:

We answer each part separately

A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude

vector quantities: the speed of a car number is the magnitude and direction is where it goes

Force, the number is the magnitude and above that applies gives direction

Scalar magnitude: how quickly the number of the speedometer of the car

Temperature, work

B) I = 15 m height to the soil and get to calculate time = 0

        y = y₀ + v₀ t - ½ g t²

as the ball is loose its initial velocity is zero

       0 = 0 +0  - ½ g t²

       t = \sqrt{2y_o/g}

       t = \sqrt{2 \ 15/ 9.8}

       t = 1.75 s

C) the velocity to the reach the floor

      v = vo - g t

      v = 0 - g t

      v = - 9.8 1.74

       v = - 17 2 m / s

The negative signt iindicates that the speed goes down

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A 0.17kg ball rolls at 0.75m/s to the right on a frictionless surface and collides with a 0.17kg ball rolling to the left at 0.6
mario62 [17]

Both momentum and kinetic energy are conserved in elastic collisions (assuming that this collision is perfectly elastic, meaning no net loss in kinetic energy)

To find the final velocity of the second ball you have to use the conversation of momentum:

*i is initial and f is final*

Δpi = Δpf

So the mass and velocity of each of the balls before and after the collision must be equal so

Let one ball be ball 1 and the other be ball 2

m₁ = 0.17kg

v₁i = 0.75 m/s

m₂ = 0.17kg

v₂i = 0.65 m/s

v₂f = 0.5

m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f

Since the mass of the balls are the same we can factor it out and get rid of the numbers below it so....

m(v₁i + v₂i) = m(v₁f + v₂f)

The masses now cancel because we factored them out on both sides so if we divide mass over to another side the value will cancel out so....

v₁i + v₂i = v₁f + v₂f

Now we want the final velocity of the second ball so we need v₂f

so...

(v₁i + v₂i) - v₁f = v₂f

Plug in the numbers now:

(0.75 + 0.65) - 0.5 = v₂f

v₂f = 0.9 m/s


8 0
3 years ago
A 12 kg block is accelerating at the rate of 9.0 m/s? while being acted on by two forces.
Aliun [14]

Answer:

See below

Explanation:

F = ma

F = 12 * 9 = 108 N

  108 N  needed      <u> add  30 N more east </u>

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1 year ago
An object has a mass of 12.8 kg and a velocity of 8.4m/s. what is the kinetic energy of the oblects
tino4ka555 [31]
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according to the law of conservation of energy what ultimatly happens to the total amount of energy in the universe?
fiasKO [112]
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8 0
3 years ago
Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.
kow [346]

Answer:

a)   # lap = 301.59 rad , b)   L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

          # turns = 1 turn (24 cm / 0.5 cm)

         # laps = 48 laps

Let's reduce to radians

        # laps = 48 laps (2 round / 1 round)

       # lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

          L₀ = 2π R

          L = # turns L₀

          L = # turns 2π R

          L = 48 2π 30

          L = 9047.79 cm

          L = 90.48 m

6 0
4 years ago
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