Answer:
2 is the numerical answer.
Explanation:
Hello there!
In this case, according to the given information and formula, it is possible for us to remember that equation for the calculation of the average kinetic energy of a gas is:
Whereas R is the universal gas constant, NA the Avogadro's number and T the temperature.
Which means that for the given ratio, we can obtain the value as follows:
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Equations of the vertical launch:
Vf = Vo - gt
y = yo + Vo*t - gt^2 / 2
Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2
=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s
The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.
Answer: 11.25 m/s
Answer:
-0.7 m/sec
Explanation:
Mass of first block = m1 =3.0 kg
Mass of second block = m2= 5.0 kg
Velocity of first block = V1= 1.2 m/s
Velocity of second block = V2 = ?
Momentum of Center of mass MVcom is sum of both blocks momentum and is given by
MVcom= m1v1+m2v2
Where
M= mass of center of mass
Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)
Putting values, we get;
0= 3×1.2+5v2
==> v2= 3.6/5= - 0.7 m/s
-ve sign indicates that block 2 is moving in opposite direction of block 1
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Efficiency = Work Output / Work Input
92% = Work Output / 100
0.92 = Work Output / 100
Work Output = 0.92 * 100
Work Output = 92 joules.
