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netineya [11]
2 years ago
8

15 beats are heard in 5 seconds when a piano key and a tuning fork of slightly higher frequency (440 Hz) are struck at the same

time. What is the frequency of the sound emitted by the piano key?
Physics
1 answer:
chubhunter [2.5K]2 years ago
6 0

Answer:

437Hz

Explanation:

if two sound waves have a little difference in frequencies f_{a}     and f_{a}       interfere, the resulting sound wave produces a beat.

Given:

f_{a}=440Hz

 f_{b}=frequency\ of\ the\ sound\ emitted by\ the\ piano\ key

f_{beat}=frequency\ of\ beat

where

f_{beat}=\frac{number\ of\ beat}{time}

f_{beat}=\frac{15beats}{5}=3Hz

f_{beat}=f_{a}-f_{b}

f_{b}=f_{a}-f_{beat} = 440Hz - 3Hz = 437Hz

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A 6 kg bowling ball moves with a speed of 3 m/s. How fast does a 7 kg bowling ball need to move so that it has the same kinetic
maw [93]

Answer: 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy.

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion.

K.E=\frac{1}{2}mv^2

m = mass of object

v= velocity of the object

K.E=\frac{1}{2}\times 6kg\times (3m/s)^2=27Joules

b) for a 7 kg bowl to have kinetic energy of 27 Joules:

27J=\frac{1}{2}\times 7kg\times v^2

v^2=7.7

v=2.8m/s

Thus 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy

4 0
2 years ago
Part A What is the resistance of a 4.4 m length of copper wire 1.3 mm n diameter? The resistivity of copper is 1.68x 10-8 Ω-m Ex
Natalija [7]

Explanation:

It is given that,

Length of the copper wire, l = 4.4 m

Diameter of copper wire, d = 1.3 mm = 0.0013 m

Radius of copper wire, r = 0.00065 m

The resistivity of the copper wire, \rho=1.68\times 10^{-8}\ \Omega-m

We need to find the resistance of the copper wire. It is given by :

R=\rho\dfrac{l}{A}

R=1.68\times 10^{-8}\ \times \dfrac{4.4\ m}{\pi (0.00065)^2}

R =0.055 ohms

So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.

7 0
3 years ago
How are different wavelengths of light affected by entering a prism?
NeX [460]
What happens in the prism stays in the prism. When the light emerges, it has the same frequency and wavelength as when it entered. The prism permanently alters nothing but the angle.

<span>Reference https://www.physicsforums.com/threads/how-does-a-prism-affect-wavelength.489768/ by caseytrimble

Sorry this probably doesn't help


</span>
3 0
3 years ago
Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
fenix001 [56]

Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

which yields:

F_{CP}=4.20 GN

Now we can find its components:

F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

F_{CPy}=-766.81 MN j

F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

7 0
3 years ago
0.054x2.33x90............
guapka [62]
<h2>0.054×2.33×90</h2><h3>=0.11582×90</h3><h3>=11.3238</h3>

please mark this answer as brainlist

5 0
2 years ago
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