All categories

3 years ago

Question is in the uploaded photo

Physics

1 answer:

ss7ja [257]3 years ago

4 0

Answer:

1) Mass of the string = 0.0267kg

Explanation:

1) In the above question, we are given:

Length of guitar string = 60cm

Wave speed = 30.0m/s

Tension = 50N

9.81 N = 1 kg

50N = 50/9.81

= 5.10kg

We have the formula

v = √(F/μ)

Where F = the tension in the string in newtons,

v = the wave speed in m/s

μ = the mass per unit length of the string in kg/m

The formula for μ is derived as

v = √(F/μ)

Square both sides

v² = [√(F/μ)]²

v² = F/μ

μ = F/v²

From the question,

μ = 40/ 30.0²

μ = 0.044kg/m

To calculate the mass of the string we use the formula

μ = Mt / Lt = total mass/total length.

Total length = 60cm

In meters = 0.60m

0.044kg/m = Mt/ 0.60m

Mt = 0.0266666667kg

Mass of the string = Approximately 0.0267kg

You might be interested in

If the photon scatters in the backward direction, what is the magnitude of the linear momentum of the electron just after the co

Strike441 [17]

Momentum is a vector quantity, and is always conserved. Whenever a collision occurs between two objects, the objects behave under the principle of conservation of momentum. Therefore, if an object moves in the direction opposite to its original direction after a collision, then this indicates that the momentum of the colliding object was greater than the object under consideration.

8 0

3 years ago

What happens to the temperature of a substance while it is changing state?

MArishka [77]

It stays constant, because it's using that energy to change state

3 0

3 years ago

Read 2 more answers

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$