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3 years ago

10

The energy supplied to a speaker is increased. What will happen to the sound the speaker produces? A) The sound will be lower pi

tch. B) The sound will become a high pitched squeal. C) The sound will be louder or greater amplitude. D) The sound will remain unchanged if the speaker is not changed.

2 answers:

Irina-Kira [14]3 years ago

7 0

Answer: The sound will be louder or greater amplitude.

Explanation: The energy input in the speaker relates directly to the intensity of the soundwave that the speaker creates, and the intensity of a soundwave is directly related to the "volume" of the sound. This is because the energy imput is used in the amplifiction proces, as more energy comes, bigger is the amplification of the sound.

So the correct option is the C: The sound will be louder or greater amplitude.

valina [46]3 years ago

6 0

Answer:

C) The sound will be louder or greater amplitude.

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A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no

Rzqust [24]

Answer:

a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

Cos 20 = Aₓ / A

sin 20 = $A_{y}$ / A

Aₓ = A cos 160 = 2 cos 160

$A_{y}$ = A sin160 = 2 sin160

Aₓ = -1,879 mi

$A_{y}$ = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

cos 2600 = Bₓ / B

sin 260 = $B_{y}$ / B

Bₓ = B cos 260

$B_{y}$ = B sin 260

Bₓ = 4 cos 260

$B_{y}$ = 4 sin 260

Bₓ = -0.6946mi

$B_{y}$ = - 3,939 mi

Third vector C = 3 mi to 15 north east

cos 15 = Cₓ / C

sin15 = $C_{y}$ / C

Cₓ = C cos 15

$C_{y}$ = C sin15

Cₓ = 3 cos 15

$C_{y}$ = 3 sin 15

Cₓ = 2,898 mi

$C_{y}$ = 0.7765 mi

Now we can find the final position of the person

X = Aₓ + Bₓ + Cₓ

X = -1.879 -0.6949 + 2.898

X = 0.3241 mi

Y = $A_{y}$ + $B_{y}$ + $C_{y}$

Y = 0.684 - 3.939 +0.7765

Y = -2.4785 mi

a) We use Pythagoras' theorem

R = √ (x2 + y2)

R = √ (0.3241 2 + (-2.4785) 2)

R = 2.4996 mi

R = 2.5 mi

b) let's use trigonometry

Tan θ = y / x

Tanθ = -2.4785 / 0.3241

θ = tan⁻¹ (-7,647)

θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

3 0

2 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago

A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

$\sum F_{x}=0$

we can see there are only two forces in x, which are the weight on x and the static friction, so:

$-W_{x}+f_{s}=0$

when solving for the static friction we get:

$f_{s}=W_{x}$

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

$W_{x}=mg sin \alpha$

we can substitute this on our sum of forces equation:

$f_{s}=mg sin \alpha$

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

$f_{s}=N\mu_{s}$

so we can substitute this on our equation:

$N\mu_{s}=mg sin \alpha$

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

8 0

3 years ago

A student believes a metal rod will gain a negative static charge when

Dennis_Churaev [7]

Answer:

no, when a plastic rod is rubbed with a duster, electrons are transferred from one material to the other. The material that gains electrons becomes negatively charged. The material that loses electrons becomes positively charged.

Explanation:

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2 years ago

Which of the following statements describes a battery?

LiRa [457]

Answer:

The negative electrode of a battery has an excess of positive charge

Explanation:

This is because in every battery, there is a negative electrode ( cathode ) and only positive charge is deposited on it.

For other statements:

Batteries donot store electric charge but they store chemical energy

Some batteries donot use metals for the flow of electrons, but some use hydrogen gas at a pressure of 1 atmosphere.

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3 years ago