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3 years ago

The loudness of a sound is the wave's _______

Physics

2 answers:

Burka [1]3 years ago

5 0

Amplitude:)

Hope this help!

levacccp [35]3 years ago

3 0

Answer:

amplitude

Explanation:

The loudness of a musical sound is a measure of the sound wave's ?

is amplitude explanation:- The loudness of a sound depends upon the amplitude.Loudness of a sound depends on the amplitude of the vibration producing that sound. Greater is the amplitude of vibration, louder is the sound produced by it. if you find this answer helpful please rate positive thank you so much.

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A cannon is fired straight up into the air. If the cannon ball comes back down to the launch point in 5 seconds, what was the ma

Nana76 [90]

Answer:

30.63 m

Explanation:

From the question given above, the following data were obtained:

Total time (T) spent by the ball in air = 5 s

Maximum height (h) =.?

Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Total time (T) spent by the ball in air = 5 s

Time (t) taken to reach the maximum height =.?

T = 2t

5 = 2t

Divide both side by 2

t = 5/2

t = 2.5 s

Thus, the time (t) taken to reach the maximum height is 2.5 s

Finally, we shall determine the maximum height reached by the ball as follow:

Time (t) taken to reach the maximum height = 2.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

h = ½gt²

h = ½ × 9.8 × 2.5²

h = 4.9 × 6.25

h = 30.625 ≈ 30.63 m

Therefore, the maximum height reached by the cannon ball is 30.63 m

3 0

3 years ago

Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the

sdas [7]

Answer:

The new distance is d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value ( $q_{B}$ =q₂ = q₂ / 5) than new distance (d = n d₀)

dat

q₁ = $q_{A}$

q₂ = $q_{B}$

r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

F = k q₁ q₂ / d₀²

F = k q₁ (q₂ / 5) / (n d₀)²

.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

5 n² = 1

n = √ 1/5

n = 0.447

The new distance is

d = 0.447 d₀

6 0

3 years ago

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Lilit [14]

Answer:

.......................

6 0

3 years ago

If a 25 kg object is moving at a velocity of 10 m/s, the object has<br> energy. Calculate it.

Paul [167]

Answer: Your answer is 1250J

Explanation:

K E = 1/2 m v 2

The mass is

m = 25 k g

The velocity is v = 10 m s − 1

So,

K E = 1 /2 x25 x 10 2^2= 1250 J

pls mark brainiest answer

4 0

3 years ago

Very lost please help.

V125BC [204]

A) The acceleration is due to gravity at any given point if you look at it vertically, so $-10 m/s^2$ .

b) $sin(25) = V_y/V$ , so $V_y = V*sin(25)$ . We use $V = V_0 + a t$ and then the final speed must be 0 because it stops at the highest point. So $0 = V_y - 10t$ . Solve for $t$ and you get $t = 32sin(25)/10 = 16sin(25)/5$

c) $Y = Y_0 + V_0t + (1/2)at^2$ , and then we plug the values: $Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2$ and we already have the time from "b)", so $Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2$ ; then we just rearrange it $Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]$ and finally $Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20$

6 0

3 years ago