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Fed [463]
2 years ago
7

QUESTION 5 When an instrument is sounded together with a turning fork of frequency 260Hz , 2 beats are heard. When the same inst

rument is sounded with a fork of frequency 256H2, , 6 beats are heard. Find the frequency of the instrument .​
Physics
1 answer:
Musya8 [376]2 years ago
4 0

Answer:

Explanation:

4126h2.

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In my opinion,I think the answer is b.You can control variables more easily by doing different things for different purposes.
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3 years ago
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Which of the following is true for a parallel circuit?
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Answer:

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3 0
2 years ago
A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i
laiz [17]

(a) Let <em>x</em> be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work <em>W</em> done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to <em>x</em> is

<em>W</em> = - (1/2 <em>kx</em> ² - 1/2 <em>k</em> (0.0400 m)²)

(note that <em>x</em> > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to <em>x</em>, so

<em>W</em> = ∆<em>K</em> = 0 - 1/2 <em>m</em> (0.835 m/s)²

Solve for <em>x</em> :

- (1/2 (160 N/m) <em>x</em> ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   <em>x</em> ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

<em>W</em> = - 1/2 <em>k</em> (0.0400 m)²

If <em>v</em> is the glider's maximum speed, then by the work-energy theorem,

<em>W</em> = ∆<em>K</em> = 1/2 <em>m</em> (0.835 m/s)² - 1/2 <em>mv</em> ²

Solve for <em>v</em> :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) <em>v</em> ²

==>   <em>v</em> ≈ 1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(<em>k</em>/<em>m</em>) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0
2 years ago
Collapse question part Part 4 (d) What is the unit vector in the direction of the spacecraft's velocity? (Express your answer in
Maksim231197 [3]

Answer:

unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

Explanation:

Given:

                            v =  (-23.2, -104.4, 46.4) m/s

Above expression describes spacecraft's velocity vector v.

Find:

Find unit vector in the direction of spacecraft velocity v.

Solution:

Step 1: Compute magnitude of velocity vector.

                            mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)

                            mag (v) = 116.58 m/s

Step 2: Compute unit vector unit (v)

                            unit (v) = vec (v) / mag (v)

                            unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58

                            unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

7 0
3 years ago
Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30
beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

J=-D\frac{\Delta C}{\Delta x}

\Delta C is the rate in concentration

\Delta xis the rate in thickness

D is the diffusion coefficient, where,

D= D_0 exp(\frac{Q_d}{RT})

Replacing D in the first law,

J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}

clearing T,

T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}

Replacing our values

T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}

T=-\frac{80000}{-138.09}

T=575.16K

4 0
3 years ago
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