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3 years ago

The______ of a sound wave is defined as the amount of energy passing through a unit area of the wave front in a unit of time

1 answer:

4 0

The intensity of a sound wave is defined as the amount of energy passing through a unit area of the wave front in unit of time.

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If you want to know how energy will move between two objects, what do you

LenaWriter [7]

Answer:

I believe it is C. Their Temps.

Explanation:

Hope my answer has helped you!

8 0

3 years ago

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

3 years ago

Yea, gonna need some help. Thanks

natulia [17]

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

$h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t$

where,

v₀ = initial speed = 110 ft/s

Therefore,

$110 = 32t\\\\t = \frac{110}{32}\\\\$

8 0

3 years ago

Name some of the mediums that sound can travel through

Brums [2.3K]

Answer:

gas, liquid, and solid

Explanation:

6 0

3 years ago

Read 2 more answers

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2010 × 103 seconds (about

vazorg [7]

Answer:

Radial acceleration of moon is $a_{r} = 2.246\times 10^{-3}$ $\frac{m}{s^{2} }$

Explanation:

Given :

Time period $T = 1.987 \times 10^{6}$ sec

Distance from center of moon to planet $r = 225 \times 10^{6}$ m

From the equation of radial acceleration,

$a_{r} = r\omega ^{2}$

Where $\omega = 2\pi f = \frac{2\pi }{T}$

So $\omega = 3.16 \times 10^{-6} \frac{rad}{s}$

Now moon's radial acceleration,

$a_{r} = 225 \times 10^{6} \times (3.16 \times 10^{-6} )^{2}$

$a_{r} = 2246.76 \times 10^{-6}$

$a_{r} = 2.246\times 10^{-3}$ $\frac{m}{s^{2} }$

7 0

3 years ago