Answer:
The number of particles in state E0 over the number of particles in state E1 will reduce
Explanation:
E0 represents the ground level state when all the particles have same energy level.
E1 represents excited state in which only a few particle reaches
E0 and E1 get further apart means that the energy difference between the two level increases.
Thus, the number of particles in state E0 over the number of particles in state E1 will reduce.
Answer:
The peak value of the electric field is 489.64 V/m
Explanation:
Given;
power of the laser, P = 1.0 mW = 1 x 10⁻³ W
Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m
Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²
The average intensity of the light = P / A
The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)
The average intensity of the light = 318.27 W/m²
The peak value of the electric field is given by;
Therefore, the peak value of the electric field is 489.64 V/m.
Answer:
D. Newton's second law
Explanation:
Newton's second law of motion states that force of an object is a product of its mass and its acceleration.
Mathematically, F= ma where m is mass and a is acceleration
So from the statement above : The acceleration of an object is proportional to the force applied to it and inversely proportional to its mass , it can be seen from the formula variation as;
F= ma -----making a the subject of the formula
a= F/ m
a= 1/m * F --------- a is inversely related to m as you can see from 1/m but directly related to F hence;
Increase in mass with the same force applied causes the body to accelerate slower where as when force increases, the body accelerates faster.
Answer:
7.468 kN
Explanation:
Here the force is given in Newton
Some of the prefixes of the SI units are
kilo = 10³
Mega = 10⁶
Giga = 10⁹
Tera = 10¹²
The number is 7468.0
Here, the only solution where the number of significant figures is kilo. If any other prefix is chosen then the significant figures will increase.
1 kilonewton = 1000 Newton
So, 7468 N = 7.468 kN
Answer:
Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is
The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is
The friction forces are computed by
Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]
Simplifying
Plugging in the values