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Alja [10]
3 years ago
6

ممكن حل بسرعة please help ​

Physics
2 answers:
aleksley [76]3 years ago
7 0
I don’t understand anything but Good luck u got it!!!!
jarptica [38.1K]3 years ago
3 0

Answer:

I could help but i dont speak or understand a word of Arabic im so sorry.

Explanation:

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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
Read 2 more answers
As you heat a steak on a grill, what kind of energy are you increasing in the steak?
katrin2010 [14]
The correct answer to the question above is Thermal Energy. As you heat a steak on a grill, the kind of energy that you are increasing is Thermal Energy. Thermal energy is the energy that come from heat, since you are you grilling the steak which means there is fire underneath, it has thermal energy. 
8 0
3 years ago
How are gravity and friction similiar
oee [108]
Gravity is the force that attracts to bodies toward each other friction is the resistance from one thing to another
8 0
3 years ago
The sun's radiation has different wavelengths. the shorter is the radiation's wavelength, the weaker is the radiation's energy.
NeX [460]
It is false. Because the amount of energy carried in the wave is inversely related to the length of the waves wavelength. To correct the statement it should be that the shorter the radiation's wavelength the stronger is the radiation's energy.
7 0
3 years ago
A kite 40 ft above the ground moves horizontally at a constant speed of 10 ft/s, with a child, holding the ball of kite string,
Lorico [155]

Answer:

 v = 27.28 m /s, θ = 63.9º

Explanation:

For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.

* We must find the horizontal speed, for this we will find the descent time and the horizontal distance

* We look for the vertical speed

At the highest point the speed is horizontal

Let's find the time it takes for the kite to reach the ground

             y = y₀ + v_{oy} t - ½ g t²

             0 =y₀ + 0 -1/2 gt²

             t = \sqrt{ \frac{2y_o}{g} }

             t = √(2 40/32)

             t = 2.5 s

to find the horizontal velocity we must know the horizontal distance, let's use trigonometry

          sin θ = y / l

          θ = sin⁻¹1 y / l

          θ = sin⁻¹ 40/50

          θ = 53.1º

therefore the horizontal distance is

          x = l cos 53.1

          x = 50cos 53.1

          x = 30 m

let's use the equation

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 30 / 2.5

          v₀ₓ = 12 m / s

we look for the vertical component of the velocity

          v_y = v_{oy} - g t

          v_y = 0 - g t

          v_y = - 9.8 2.5

          v_y = -24.5 m / s

the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign

We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus

          v = \sqrt{v_x^2 + v_y^2}

          v = \sqrt{12^2 + 24.5^2}

          v = 27.28 m /s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ v_y / vₓ

          θ = tan⁻¹ 24.5 / 12

          θ = 63.9º

4 0
3 years ago
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