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Tju [1.3M]
3 years ago
8

Two people are talking at a distance of 3.0 m from where you are, and

Physics
1 answer:
tigry1 [53]3 years ago
5 0

Answer:

The sound intensity that the other student measure, I₂ is expected to be;

A) 6.2 × 10⁻⁸ W/m²

Explanation:

The measure of sound intensity is given by the following formula;

I = \dfrac{P}{4 \cdot \pi \cdot R^2}

\dfrac{I_2}{I_1} = \dfrac{R_1^2}{R_2^2}

Where;

I = The intensity

R = The radius

P = The power of the sound

Whereby we have;

The distance of the two people talking, R₁ = 3.0 m

The measure of the sound intensity, I₁ = 1.1 × 10⁻⁷ W/m² (from an online source)

The distance of the other student from the two people talking, R₂ = 4.0 m

Therefore, the estimate of the sound intensity, I₂, is given as follows;

{I_2} = \dfrac{R_1^2}{R_2^2} \times {I_1}

{I_2} = \dfrac{(3.0 \, m)^2}{(4.0 \, m)^2} \times 1.1 \times 10^{-7} \ W/m^2 = 6.1875 \times 10^{-8} \ W/m^2

I₂ = 6.1875 × 10⁻⁸ W/m²

∴ The sound intensity that the other student measure, I₂ ≈ 6.2 × 10⁻⁸ W/m²

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elena55 [62]
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\alpha =  \frac{\omega_f - \omega_i}{\Delta t}
where
\omega_f is the final angular speed of the object
\omega_i is its initial angular speed
\Delta t is the time taken to accelerate

For the wheel in our problem, \omega_f=11.1 rad/s, \omega_i = 0 and \Delta t=2.99 s, so its angular acceleration is
\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2
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3 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
IRINA_888 [86]

Answer:

Time : <u>7.96 s</u>

Distance Traveled : <u>357.8 m</u>  

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s    (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s    (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s    (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

<u>t₂ = 7.96 s</u>

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

<u>S = 357.8 m</u>

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Instantaneous center:

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From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

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To know more about velocity, refer: brainly.com/question/12413963

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1 horsepower = 746 watts = 746 joule/second

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=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
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enyata [817]

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d= (vi+vf)/2 x t </span>

where t = time

vi is the initial velocity = 23 m/s

<span>vf is the final velocity  = 0 m/s</span>

and d is the distance covered = 85m

85 = (23 + 0)/2 x t

t = 85/11.5 = 7.39s

Now use this equation to find acceleration:

a= (vf-vi)/t <span>

a = 0 – 23/ 7.39</span>

a = -3.112 m/s²

<span>so, the acceleration is 3.112 m/s</span>²<span>


</span>

6 0
3 years ago
Read 2 more answers
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