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3 years ago

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Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc

y of 5.0 Hz. One of the violinists measures with an electronic tuner and finds her A string is at 462 Hz. What are the possible frequencies for the A string of the other violinist?(multiple answers)
462 Hz
452 Hz
457 Hz
467 Hz
472 Hz

1 answer:

kaheart [24]3 years ago

4 0

Answer:

The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.

(3) and (4) is correct option.

Explanation:

Given that,

Beat frequency f = 5.0 Hz

Frequency f'= 462 Hz

We need to calculate the possible frequencies for the A string of the other violinist

Using formula of frequency

$f'=f_{1}-f$ ...(I)

$f'=f_{1}+f$ ...(II)

Where, f= beat frequency

f₁ = frequency

Put the value in both equations

$f'=462-5=457\ Hz$

$f'=462+5=467\ Hz$

Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.

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A mass of 5kg accelerates at 3m/s/s, how much force was put on it?

goldfiish [28.3K]

Answer:

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A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l

mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be $\who_w = 1000 kg/m^3$

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

$F_s + F_b = W$

Where $F_s = kx$ N is the spring force, $F_b = W_w = m_wg = \rho_w V_s g$ is the buoyancy force, which equals to the weight $W_w$ of the water displaced by the submerged portion of the cylinder, which is the product of water density $\rho_w$ , submerged volume $V_s$ and gravitational constant g. W = mg is the weight of the metal cylinder.

$kx + \rho_w V_s g = mg$

The submerged volume would be the product of cross-section area and the submerged length x

$V_s = Ax = \pi(d/2)^2x$

Plug that into our force equation and we have

$kx + \rho_w \pi(d/2)^2x g = mg$

$x(k + \rho_w g \pi d^2/4) = mg$

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2 years ago

A 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

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Their velocity afterward is v=3.467 m/s

Explanation:

Given:

Mass of the first football player= 91.5 kg

Initial velocity of the football player 3.73 m/s

Mass of second football player=63.56 kg

Initial velocity of the second football player=3.09 m/s

To find:

Final velocity of both players=?

Solution:

According to the law of conservation of momentum,

Initial momentum =final momentum

mathematically represented as

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...........................(1)

where

$u_1$ =intial velocity of the football player

$u_2$ = inital velocity of second football player

$v_1$ =finall velocity of the first football player

$v_2$ =final velocity of second football player

after tackling , both the football players moves with the same velocity,

so $v_1=v_2=v$

Hence equation (1) becomes

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

now substituting the values,

$v=\frac{(91.5\times+3.73)+(63.5\times3.09)}{(91.5+63.5)}$ $v=\frac{(341.29+196.210)}{155}$

$v=\frac{537.5}{155}$

v=3.467 m/s

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