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zavuch27 [327]
2 years ago
15

When a 440-Hz tuning fork and a piano key are struck together, five beats are heard. If the pitch of the note on the piano is lo

wer than the tuning fork, what is the frequency of the note?
Physics
2 answers:
vovangra [49]2 years ago
7 0
The frequency would also be lower
guapka [62]2 years ago
6 0
440Hz < n2
n2 - 440 = 5
n2 = 445hz
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evaluate the numerical value of the vertical velocity of the car at time t=0.25 s using the expression from part d, where y0=0.7
likoan [24]

Given :

Displacement , y = 0.75 m .

Angular acceleration , \alpha=0.95\ s^{-2} .

Initial angular velocity , \omega_o=6.3\ s^{-1} .

To Find :

The value of vertical velocity after time t = 0.25 s .

Solution :

By equation of circular motion is given by :

\omega=\omega_o+\alpha t

Putting all given values we get :

\omega=6.3+0.95\times 0.25\\\\\omega= $$6.5375\ s^{-1}

Now , vertical velocity is given by :

v=y\omega\\\\v=0.75\times 6.5375\ m/s\\\\v=4.90\ m/s

Therefore , the numerical value of the vertical velocity of the car at time t=0.25 s is 4.90 m/s .

Hence , this is the required solution .

8 0
2 years ago
. Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the
vichka [17]
Transverse wave as the wave is going up and down no compressions
3 0
2 years ago
The planet Neptune orbits the Sun. Its orbital radius is 30.130.130, point, 1 astronomical units (\text{AU})(AU)left parenthesis
lord [1]

Answer:

The distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>

Explanation:

As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is <em>equivalent</em> to the distance it travels around the Sun in a single orbit.

<em>The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)</em>

Where,

<em>R = Orbital radius (in this case) = 30.1 AU</em>

<em />

Plug the value of R in the equation (A):

<em>(A) => The circumference of the circle = 2*π*(30.1)</em>

<em> The circumference of the circle = </em><em>60.2π</em>

Therefore, the distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>

5 0
2 years ago
A coworker did not clean his work area before going home this could cause an accident so you quickly clean up the next day you s
defon

Answer:

THE FIRST ONE YOU SHOULD TELL HIM AND THE LAST ONE YOU SHOUDENT DO BECAUSE HE WILL DO IT AGAIN AND EXPECT OTHERS TO CLEAN UP AFTER HIM

Explanation:

5 0
1 year ago
Read 2 more answers
Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi
aleksklad [387]
1) The buoyant force acting on an object immersed in a fluid is:
B=d_f V_d g
where d_f is the density of the fluid, V_d is the volume of displaced fluid, and g=9.81~m/s^2 is the gravitational acceleration.

2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3. Using the relationship between density, volume and mass, we find
V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3

3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is d_f = 1~Kg/m^3, so we have
B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N

4) The weight of 1 Kg of titanium is instead:
W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N
So, the buoyant force is negligible compared to the weight.
7 0
1 year ago
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