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3 years ago

Which is a property of bases?

Physics

2 answers:

vesna_86 [32]3 years ago

4 0

Answer:

C. Slippery feel

Explanation:

padilas [110]3 years ago

3 0

Slippery feel would be the correct answer (:

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There is evidence that elephants communicate via infrasound, generating rumbling vocalizations as low as 14 Hz that can travel u

amid [387]

Answer:

$L = 96.2 dB$

Explanation:

As we know that the level of intensity is given as

$\beta = 10Log(\frac{I}{I_o})$

here we know

$\beta = 104 dB$

$104 = 10Log(\frac{I}{10^{-12}})$

$I = 0.025 W/m^2$

now the sound is travelling in all possible directions so we have

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{0.025}{I_2} = \frac{10^2}{4.1^2}$

$I_2 = 4.2 \times 10^{-3} W/m^2$

now for level of sound we have

$L = 10Log(\frac{4.2 \times 10^{-3}}{10^{-12}})$

$L = 96.2 dB$

5 0

3 years ago

Why is people asking for segges this is so annoying how do i block them help mee

NikAS [45]

Answer:

woah

Explanation:

report

6 0

3 years ago

The container was lifted a height of 14 m

Alika [10]

Work done = Force X Distance

3 430 000J = Force X 14m

Force = 3 430 000J / 14m
= 245 000 N

Hope this helps!

8 0

2 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is: