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Butoxors [25]
3 years ago
6

During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h

as an angular speed of 12.0 rad/s. If the distance between the top of the racket and the shoulder is 1.30 m, find the magnitude of the total acceleration of the top of the racket.
Physics
1 answer:
QveST [7]3 years ago
3 0

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

a

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

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NemiM [27]

Answer:

- the coating’s index of refraction is 1.25

- the required thickness is 104.1667 nm

Explanation:

Given the data in the question;

Thickness of coating t = 100 nm

wavelength λ = 500nm

we know that refractive index is;

t = λ/4n

make n, the subject of formula

t4n = λ

n = λ / 4t

we substitute

n = 500 / ( 4 × 100 )

n = 500 / 400

n = 1.25

Therefore, the coating’s index of refraction is 1.25

2)

given that;

Index of refraction of the coating; n = 1.20

λ = 500 nm

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we substitute

t = 500 / ( 4 × 1.2 )

t = 500 / 4.8

t = 104.1667 nm

Therefore, the required thickness is 104.1667 nm

4 0
3 years ago
Under what condition is the instantaneous acceleration of a moving body equal to its average acceleration over time?
Rzqust [24]
If the acceleration is constant (negative or positive) the instantaneous acceleration cannot be

Average acceleration: [final velocity - initial velocity ] /Δ time

Instantaneous acceleration = d V / dt =slope of the velocity vs t graph

If acceleration is increasing, the slope of the curve at one moment will be higher than the average acceleration.

If acceleration is decreasing, the slope of the curve at one moment will be lower than the average acceleration.

If acceleration is constant, the acceleration at any moment is the same, then only at constant accelerations, the instantaneuos acceleration is the same than the average acceleration.

Constant zero acceleration is a particular case of constant acceleration, so at constant zero acceleration  the instantaneous accelerations is the same than the average acceleration: zero. But, it is not true that only at zero acceleration the instantaneous acceleration is equal than the average acceleration.

That is why the only true option and the answer  is the option D. only at constant accelerations.
3 0
3 years ago
12. An object accelerates 16.3 m/s2 when a force of 4.6 newtons is applied to it.
ladessa [460]
The acceleration of body is given 16.3m/s2 and the force is given 4.6 N then
We know,
Force=mass*acceleration
Then,
Mass=force/acceleration
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Four students are each pushing on one side of a box. Mara is pushing to the west, Cindi is pushing to the east, Chris is pushing
jenyasd209 [6]

Answer:

cindi

Explanation:

cindi's work done is larger than all the other students combined

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