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3 years ago

5

The decibel level of an orchestra is 90 db, and the violin section achieves a level of 80 dB. How does the sound intensity from

the full orchestra compare to that from the violin section alone?

1 answer:

skad [1K]3 years ago

4 0

Answer:

The difference in the decibel corresponses to a constant difference in the loudness perceived.

The refore the sound intensity from the orchestra is like 100 times that of the violin.

Explanation:

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A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m

Svetach [21]

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

= 67(9.41 - 0)

= 67 x 9.41

= 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

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3 years ago

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Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it

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<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>

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3 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

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2 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

2 years ago