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3 years ago

Forces normal to a particle's displacement do no work.

Physics

1 answer:

snow_tiger [21]3 years ago

3 0

Answer:

Explanation:

The work done is defined as the product of force applied in the direction of displacement and the displacement.

W = F x d x Cosθ

where, F is the force applied, d be the displacement and θ be the angle between the displacement and force.

For the normal forces, the angle between the displacement and the force applied is 90 degree, and the value of Cos 90 is zero, so the work done is zero.

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In addition to a reference point you also need distance and __________ to describe location

Lana71 [14]

displacement

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2 years ago

· A car moves at 12 m/s and coasts up a hill with a

Zigmanuir [339]

Explanation:

Given that,

Initial speed of a car, u = 12 m/s

Aceleration, a = -1.6 m/s²

(a) The displacement of car after 6 seconds is :

$d=ut+\dfrac{1}{2}at^2\\\\d=12\times 6+\dfrac{1}{2}\times (-1.6)\times 6^2\\\\d=43.2\ m$

(b) The displacement of the car after 9 seconds is :

$d=ut+\dfrac{1}{2}at^2\\\\d=12\times 9+\dfrac{1}{2}\times (-1.6)\times 9^2\\\\d=43.2\ m$

Hence, this is the required solution.

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3 years ago

Select the correct answer from the drop-down menu. anne applies a force on a toy car and makes it move forward. what can be said

nalin [4]

Answer:

The forces could be gravity, friction between the car and the ground, the force Katie is applying and the normal reaction.

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1 year ago

Rahim is watching his favorite football team on television. In order to work, the television must be

kvasek [131]

Answer:

it would be the cord will have the most energy at the moment.

Explanation:

...

3 0

2 years ago

A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.

Georgia [21]

Answer:

The wavelength of the light is 562.5 nm

Solution:

As per the question:

Order, n = 1

Slit separation, d = $2.5\times 10^{- 6}\ m$

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

$n\lambda = dsin\theta$

n = 1

$\lambda = dsin\theta$ (1)

Also,

For very small angle, $\theta$ :

$sin\theta$ ≈ $tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225$

Using the above value in eqn (1):

$\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm$

3 0

2 years ago