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2 years ago

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How do you find the coordinates of centre of mass of multiple particle system ?

Physics

1 answer:

scZoUnD [109]2 years ago

8 0

By dividing the moments around the y-axis and around the x-axis,

respectively, by the total mass, the x- and y-coordinates of the centre of

mass may be determined.

formula's:

$y.cm=\left[\begin{array}{ccc}\frac{m1y1+m2y2+...}{m1+m2+...}\end{array}\right]$

$x.cm=\left[\begin{array}{ccc}\frac{m1x1+m2x2+....}{m1+m2+.....}\end{array}\right]$

centre of two mass particle system: $\left[\begin{array}{ccc}(m1+m2) r.cm =(m1 )r1+(m2) r2\end{array}\right]$

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A 0.5 kg stone is raised from 1m to 2m height from the ground. what is the change in potential energy of the stone?

Usimov [2.4K]

Given: The mass of stone (m) = 0.5 kg

Raised from heights (h₁) = 1.0 m to (h₂) = 2.0 m

Acceleration due to gravity (g) = 9.8 m/s²

To find: The change in potential energy of the stone

Formula: The potential energy (P) = mgh

where, all alphabets are in their usual meanings.

Now, we shall calculate the change in potential energy of the stone

Δ P = P₂ - P₁ = mg (h₂ - h₁)

or, = 0.5 kg ×9.8 m/s² ×(2.0 m - 1.0 m)

or, = 4.9 J

Hence, the required change in the potential energy of the stone will be 4.9 J

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3 years ago

1. A rocket is fired vertically from the launch pad with a

zavuch27 [327]

Answer:

Max height= 36000 meters

Total Time = 120 seconds

Explanation:

0 = U - at

U = at

U= 20*60

U= 1200 m/s

MAX altitude would be

(U²Sin²tita)/2g

Max height= 1200² *( SIN90)²/(2*20)

Time of FLIGHT

2 * 1200/20

2400/20

120 sec onds

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3 years ago

a 700k/g race car slowed down from 30m/s to 15m/s what is the impulse and what was the average force applied by the brakes

posledela

So impulse is a change in momentum.

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I dont think you will be able to find the average force with the given info because you need to know the time it takes for the car to slow down.

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3 years ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

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3 years ago

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