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Llana [10]
2 years ago
8

A band director instructs students to play a pitch louder. How will the sound wave change when the band plays the same pitch lou

der?
The speed of the sound wave will increase.
The frequency of the sound wave will increase.
The wavelength of the sound wave will increase.
The amplitude of the sound wave will increase.

Physics
2 answers:
Fofino [41]2 years ago
6 0

Explanation :

Loudness is defined as how our ear perceives the sound. It is basically related to the intensity of a sound wave.

When the same pitch is played louder, its amplitude changes. Amplitude is the maximum displacement covered by a wave.

This is the only difference between the loudness to that of shrillness. The amplitude of the loud sound is more as compared to that of shrill sound.

So, the correct option is (d).

seropon [69]2 years ago
3 0
The amplitude will increse

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How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 3236
Korolek [52]

In determining the number of significant figures in a given number, there are three rules to always remember / follow:

First: All integers except zero are always significant.

<span>Second: Any zeros located between  non zeroes are always significant.</span>

Third: A zero located after a non zero in a decimal is always significant whether it is before or after the decimal

 

Therefore using this rule, the number of significant digits in the given numbers are:

(a) 214 = 3

(b) 81.60 = 4

(c) 7.03 = 3

(d) 0.03 = 1

(e) 0.0086 = 2

(f) 3236 = 4

(g) 8700 = 2

4 0
2 years ago
An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and
torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

 

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

 

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552  ; from this we get AC = 172 km (3 significant figures)

 

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

 

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

 

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

5 0
1 year ago
8. Rubbing a plastic bag and a balloon with a cloth gives both objects a net negative charge. The balloon's
Dafna1 [17]

Answer:

0.214 m

Explanation:

In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.

Therefore, we can write:

k\frac{q_1 q_2}{r^2}=mg

where

k is the Coulomb constant

q_1=-1\cdot 10^{-10}C is the charge on the balloon

q_2=-1\cdot 10^{-5} C is the charge on the bag

r is the separation betwen the bag and the balloon

m=0.02 g=2\cdot 10^{-5} kg is the mass of the bag

g=9.8 m/s^2 is the acceleration due to gravity

Solving for r, we find the distance at which the bag must be held:

r=\sqrt{\frac{kq_1 q_2}{mg}}=\sqrt{\frac{(9\cdot 10^9)(-1\cdot 10^{-10})(-1\cdot 10^{-5})}{(2\cdot 10^{-5})(9.8)}}=0.214 m

5 0
1 year ago
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
mote1985 [20]

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

6 0
1 year ago
A 385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of 0.0125 kg ⋅ m2 and a radius of 1
dimaraw [331]

Answer:

the tension in the string is 5.59 N

Explanation:

Here ,

m_1 = 0.385 Kg

m_2 = 0.710 Kg

Using second law of motion ,

a = F_net / effective mass

a = (0.710- 0.385)×9.8/(0.710 + 0.385 + 0.0125/0.15^2)

a = 1.93 m/s^2

Now , let tension be T ,

then,

mg-T=ma

0.710×g - T = 0.710×1.93

T = 5.59 N

the tension in the string is 5.59 N

7 0
1 year ago
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