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BARSIC [14]
3 years ago
14

14. How are position and speed useful for describing the motion of an object?

Physics
1 answer:
lisabon 2012 [21]3 years ago
5 0

Answer:

The Position and speed of an object is very important to describe the motion of a moving body as it directly affects the rate at which the moving body approaches the given point.

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Using a simple machine, a student is able to lift a 500N weight by applying only 100N.
pishuonlain [190]
Hey there!

a \ student  \ is \ able \ to \ lift \ a \ 500(n) \\ weight \ by \ applying \ only  \ 100(n) \\ \\ so, \ this \ info \ here, \ we \ simply \ divide \ by \\ how \ much \ this \ kid \ lifted, \ by \ the \ weight \ he/she \\ \ is \ applying. \\ \\   \left[\begin{array}{ccc}\boxed{\boxed{500(n)/100(n)}}\end{array}\right] \\ \\ and \ from \ this,\ your \ answer \ would \ \\ conclude \ to \ be \ (5) \\ \\ \boxed{5} \ would \ be \ your \ answer!

Hope this helps you!
7 0
3 years ago
Need Help !
Bad White [126]

Answer: 5.96m/s

Explanation:

Given the following :

Mass of car (m) = 1500kg

Velocity (V) = 5.25m/s

Forward force of engine = 1250N

Diatance moved = 4.8m

Final Velocity =?

Final kinetic energy = Initial kinetic energy + work done by engine

Initial kinetic energy = 0.5 × mass × velocity^2

Initial kinetic energy = 0.5 × 1500 × 5.25^2

Initial kinetic energy = 20671.875 J

Work done by engine = Force × distance

Work done by engine = 1250 × 4.8 = 6000J

Final kinetic energy = (20671.875 + 6000) J

= 26671.875 J

From kinetic energy = 0.5mv^2

26671.875 = 1/2 × 1500 × v^2

53343.75 = 1500v^2

v^2 = 35.5625

v = sqrt(35.5625)

v = 5.96m/s

3 0
3 years ago
Can someone please help me with this
pentagon [3]
The key principle is that crank length, just like frame size, should be proportional to the rider height and then modified to what fits the individual. There are 4 charts, two for the upright position and two for the aero position, depending upon how you race.
4 0
2 years ago
For the image of the overhead projector to be in focus, the distance from the projector lens to the image, <img src="https://tex
rjkz [21]
Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet

convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches
 
1/f = 1/96 inches + 1/4 inches

Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96

to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f

The focal length of the project lens is 3.84 inches 

4 0
3 years ago
A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
My name is Ann [436]

Answer:

a)   C = 4,012 10⁻¹⁴ F, b)  Q = 1.6 10⁻¹¹ C , c)   U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

       C = k e₀ A / d

Let's calculate

       C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

       C = 4,012 10⁻¹⁴ F

b) let's look  the charge

        C = Q / ΔV

         Q = C ΔV

         Q = 4,012 10⁻¹⁴ 400

         Q = 1.6 10⁻¹¹ C

c) The stored energy

        U = ½ C ΔV²

        U = ½ 4,012 10⁻¹⁴  400²

        U = 3.21 10⁻¹¹ J

4 0
3 years ago
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