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2 years ago

13

Let's say you have two tuning forks which are supposed to produce the same frequency, 512 Hz. One is of good quality, but the ot

her is cheaply made and vibrates at 510 Hz. If you bonk them both how many beats per second will you hear

1 answer:

solong [7]2 years ago

7 0

Answer:

= 2 beats per seconds

Explanation:

From |f -f'| = modulus of the difference between the frequency given.

f = 510Hz and f' = 512Hz

Difference between the frequency will give us the number of beat per seconds.

i.e 2 beats per seconds

These also shows how to get the period of the tuning forks.

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A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical heig

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Answer:

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I = 2/5 m R^2 inertia of solid sphere

1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

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v = 16.3 m/s

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1 year ago

What is the pressure transmitted in the liquid on a hydraulic pump where an elephant with a weight of 40 000 N is placed on top

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What is the pressure transmitted in the liquid on a hydraulic pump where an elephant with a weight of 40 000 N is placed on top of the large piston with an area of 40 m2. The small piston area is 4 m2.

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2 years ago

Difference between velocity ratio and efficiency

Naddik [55]

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2 years ago

The addition of 9.0×105 J is required to convert a block of ice at -10 ∘C to water at 11 ∘C. i need help this is due in less tha

Law Incorporation [45]

The mass of the block of ice is 2.24 kg

Explanation:

The amount of heat needed for the whole process consists of three different amounts of heat:

$Q_1$ : the amount of heat needed to raise the temperature of the block of ice from $-10^{\circ}C$ to $0^{\circ}C$

$Q_2$ : the amount of heat needed to melt the block of ice at melting point

$Q_3$ : the amount of heat needed to raise the temperature of the water from $0^{\circ}C$ to $11^{\circ}C$

The total amount of heat needed can be written as

$Q=Q_1+Q_2+Q_3=mC_i\Delta T_1 + m\lambda_f + mC_w\Delta T_3$

where we have:

$Q=9.0 \cdot 10^5 J$ (total amount of heat required)

m is the mass of the block of ice

$C_i = 2108 J/kg^{\circ}C$ is the specific heat of ice

$\lambda_f=3.34\cdot 10^5 J/kg$ is the latent heat of fusion of ice

$C_w=4186 J/kg^{\circ}C$ is the specific heat capacity of water

$\Delta T_1 = 0-(-10)=10^{\circ}C$ is the change in temperature in the 1st process

$\Delta T_3 = 11-0=11^{\circ}C$ is the change in temperature in the 3rd process

Solving the equation for m, we find the mass of the block of ice:

$m=\frac{Q}{C_i\Delta T_1 + \lambda_f+C_w\Delta T_3}=\frac{9.0\cdot 10^5}{(2108)(10)+3.34\cdot 10^5+(4186)(11)}=2.24 kg$

Learn more about specific heat capacity:

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2 years ago

Absolute pressure in tank is P1 = 260 kPa and local ambient absolute pressure is P2 =100 kPa. If liquid density in pipe is 13600

Pavlova-9 [17]

Answer:

1.176m

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Net pressure=260-100=160kPa

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h= $\frac{160000}{136000}$

h=1.176m

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2 years ago