The kinetic molecular theory of gases predicts pressure to increase as the temperature of a gas increases because

A block is at rest. The coefficients of static and kinetic friction are s = 0.81 and k = 0.69, respectively. The acceleration of

lys-0071 [83]

Answer:

Explanation:

a ) The angle required = angle of repose = θ

Tanθ = .81

θ = 39⁰

b ) when angle of incline θ = 44

Net force on the block = mg sinθ - μ mg cosθ where μ is coefficient of kinetic friction

acceleration = gsinθ - μ g cosθ

= 9.8 ( sin44 - μ cos44 )

= 9.8 ( .695 - .69 x .72 )

= 9.8 ( .695 - .497 )

= 1.94 m /s²

7 0

3 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:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

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

3 0

3 years ago

Suppose a 4.90 m diameter telescope were constructed on the moon, where the absence of atmospheric distortion would permit excel

hichkok12 [17]

The problem is asking the maximum separation between the telescope in the moon and mars. To tell you frankly, I have no idea how to calculate this by I research the answer which I will give to you and the clue that I just know is that the distance should be in kilometers and the telescopes diameter could affect the minimum separation between the two object. With that the answer would be 422,700 km

5 0

3 years ago

A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4

zaharov [31]

We will apply the equation:
2as = v² - u²
v = √(2as + u²)
v = √(2 x 4 x 400 + 13²)
v = 58 m/s

5 0

3 years ago

Read 2 more answers

Electric powerlines in electric Post hang very low in hot summer? Why?

irina1246 [14]

Electric power lines in the electric posts hang very low during the summer to prevent overheating. They’re often overloaded with high energy consumption and overheating could cause destruction of conductors. The high temperature could melt or expand the copper and aluminum materials. To protect public safety and guarantee an efficient energy transmission, the power lines are lowered to decrease temperature.

5 0

4 years ago

Read 2 more answers