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dsp73
3 years ago
9

Give an example of a situation in which you would describe an object's position in

Physics
1 answer:
notka56 [123]3 years ago
8 0
Incomplete question.  Full text is:

"<span>Give an example of a situation in which you would describe an object's position in (a) one-dimension coordinates (b) two-dimension coordinates (c) three-dimension coordinates"

Solution
(a) One dimension example: a man walking along a metal plank. We just need to specify one coordinate, the distance from the beginning of the plank.

(b) Two-dimension example: a ball moving on a circle. In this case, we need two coordinates: (x,y) to specify the position of the ball at every instant, since it is moving on a 2-D plane.

(c) The position of an airplane in the air: in this case we need 3 coordinates, the height, the latitude and the longitude of the airplane.</span>
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A child with a weight of 230 N swings on a playground swing attached to 1.90 m long chains. What is the gravitational potential
nlexa [21]

Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

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Gravitational Potential Energy is given as:

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m = mass

h = height above the ground

W = weight

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3 years ago
A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are
slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

Explanation:

Given that,

Mass m = 1.81\times10^{-3}\ kg

Velocity v = (3.00\times10^{4}\ m/s)j

Charge q = 1.22\times10^{-8}\ C

Magnetic field B= (1.63\hat{i}+0.980\hat{j})\ T

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

F = ma=q(v\times B)

a =\dfrac{q(v\times B)}{m}

We need to calculate the value of v\times B

v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})

v\times B=4.89\times10^{4}

Now, put the all values into the acceleration 's formula

a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}

a= -0.3296\ \hat{k}\ m/s^2

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

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3 years ago
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