Help me please!!!!!!!!!!!!!!!

You have 0.5 l of air at a pressure of 203 kpa and -70°c in a rigid, sealed contaainer. what is the absolute tempertaure of the

ANTONII [103]

From the information given and if the question is complete then;
Absolute temperature is the temperature in Kelvin
To convert degree Celsius to kelvin we normally add 273
that is Kelvin = deg Celsius + 273
Thus since we have been given that the air was at -70 degrees celcius;
then; - 70° C + 273 = 203 K
Therefore; the absolute temperature is 203 K

5 0

2 years ago

The density of water is 1.00 g/cm3. What is its density in kg/m3?

r-ruslan [8.4K]

1 g = 1 ÷ 1000 kg
= 0.001 kg

1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³

1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³

The density is 1000 kg/m³.

3 0

3 years ago

A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.

Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

$\omega =\frac{2\pi \times 469}{60}=49.11 rad/s$

t=16 s

Angular deceleration in 16 s

$\omega =\omega _0+\alpha \cdot t$

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

Change in kinetic energy =Work done

Change in kinetic Energy $=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}$

$\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J$

(a)Work done =50.79 kJ

(b)Average Power

$P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW$

7 0

3 years ago

Please help me find the answer

Hoochie [10]

Answer:

26b) 66.7%

27) 500 N

Explanation:

26.a) In a two pulley system, the load is attached to one of the pulleys. The other pulley is attached to a fixed surface, as well as one end of the rope. The other end of the rope goes around moving pulley, then around the fixed pulley.

26.b) Mechanical advantage is the ratio between the forces:

MA = load force / effort force

Efficiency is the ratio between the work:

e = work done on load / work done by effort

Work is force times distance.

e = (F load × d load) / (F effort × d effort)

Rearranging:

e = (F load / F effort) × (d load / d effort)

e = MA × (d load / d effort)

In a two pulley system, the load moves half the distance of the effort. So the efficiency is:

e = (4/3) × (1/2)

e = 2/3