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Alja [10]
3 years ago
7

Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no

beats will be heard. The more experienced player plays a note with a frequency of 520 Hz. When the second musician begins to play, the note sounds lower and they can hear 20 beats in 4.0 s. With what frequency is the second trumpeter playing?
Physics
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

The second trumpeter will be playing at frequency = 515 Hz

Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s. 

Beat frequency = 20/4 = 5 Hz

Beat frequency = F2 - F1

5 = 520 - F1

F1 = 520 - 5

F1 = 515 Hz

Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

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A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o
maxonik [38]
For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
v = y*f = 1.5*220 = 330 m/s
4 0
3 years ago
In a darkened room, a burning candle is placed 1.84 m from white wall. A lens is placed between candle and wall at a location th
KATRIN_1 [288]

Answer:

82.4 cm

Explanation:

The object and screen are kept fixed ie the distance between them is fixed and by displacing lens between them images are formed on the screen . In the first case let u be the object distance and v be the image distance

then ,

u + v = 184 cm

In the second case of image formation , v becomes u and u becomes v only then image formation in the second case is possible.

The difference between two object distance ie(  v - u ) is the distance by which lens is moved so

v - u = 82.4 cm

7 0
3 years ago
Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive or negative: (a)
Aleks04 [339]

Answer:

a) As the chicken is still, the displacement is zero, which implies that the work is zero.

b) as the person is still there is no displacement therefore the work is zero

c) Lagraua applies a vertical force and the displacement is vertical, therefore the Work is positive

d) the force of gravity is directed downwards and the displacement is upwards, therefore the angle between it is 180º and the 180º fly is -1. Consequently the lock is negative

e) when the person meticulously feels the upward force and the displacement is downward, therefore the work is negative

Explanation:

Work is defined by the expression

        W = F. r

bold letters indicate vectors, we can write this expression as a module

        W= F r cos θ

where is at the angle between force and displacement.

Let's apply this expression to the different cases

a) As the chicken is still, the displacement is zero, which implies that the work is zero.

b) as the person is still there is no displacement therefore the work is zero

c) Lagraua applies a vertical force and the displacement is vertical, therefore the Work is positive

d) the force of gravity is directed downwards and the displacement is upwards, therefore the angle between it is 180º and the 180º fly is -1. Consequently the lock is negative

e) when the person meticulously feels the upward force and the displacement is downward, therefore the work is negative

7 0
3 years ago
The normal formula to find force is F = m* a. What kind of math do you need to do
Oduvanchick [21]
To find the Mass of an object, you need to apply division.
Since Resultant Force = Mass X Acceleration

To find mass,
Mass = Force / Acceleration
3 0
3 years ago
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z
sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0
3 years ago
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