Answer:
0.002833 sec
Explanation:
Speed of light in vacuum is
Given distance = 850 km = 850×1000=850000 m
We have to calculate the time that light take to travel the distance 850 km
Time
So the time taken by light to travel 850 km is 0.002833 sec
An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on of p
1 answer:
The given question is incomplete. The complete question is as follows.
An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on 1 of 36 possible sites for adsorption. Calculate the entropy of this system.
Explanation:
It is known that Boltzmann formula of entropy is as follows.
s = k ln W
where, k = Boltzmann constant
W = number of energetically equivalent possible microstates or configuration of the system
In the given case, W = 36. Now, we will put the given values into the above formula as follows.
s = k ln W
=
=
Thus, we can conclude that the entropy of this system is .
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Answer:
So airplane will be 1324.9453 m apart after 2.9 hour
Explanation:
So if we draw the vectors of a 2d graph we see that the difference in angles is = 83 - 44.3 =
Distance traveled by first plane = 730×2.9 = 2117 m
And distance traveled by second plane = 590×2.9 = 1711 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.
Using the law of cosine, representing the distance between the planes, we see that:
d = 1324.9453 m
Given data
The angle of inclination of the plane is theta = 30 degree
The applied force in the inclined plane is F = 94 N
The distance moved in the inclined plane is d = 2.30 m
The coefficient of kinetic friction is u_k = 0.280
The free-body diagram of the above configuration is shown below:
Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.
(a)
The expression for the work done by the pushing force is given as:
Substitute the value in the above equation.
Thus, the work done by the pushing force is 216.2 J.
(b)
The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.
The expression for the normal reaction force is given as:
The expression for the mass of the box is given as:
Substitute the value in the above equation.
Thus, the mass of the box is 12.9 kg.
