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astra-53 [7]
3 years ago
9

A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1

.80 m travels along the string.What is the maximum velocity of a particle on the string?
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

\mu = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s

Frequency is given by

f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz

Angular frequency is given by

\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s

Maximum velocity of a particle is given by

v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s

The maximum velocity of a particle on the string is 8.89288275 m/s

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Hope this helps. Have a nice day.
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50 kg of water at 75o C is cooled to 25o C. How much heat was given off?
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2 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
3 years ago
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