Emotional Intelligence is our own abilities of being able to identify and regulate our own emotions. This is used by individuals in the real world in order to regulate crying in front of everyone for example. Calming ourselves down by breathing is another good example of regulating emotion which plays into emotional intelligence.
Hope this helps mate :D
Answer:
Solution
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(a) The labelled diagram is shown.
(b) The refractive index of diamond is 2.42. Refractive index of diamond is the ratio of the speed of light in air to the speed of light in diamond.i.e.,
μ=
Speedoflightindiamond
Speedoflightinair
and, the ratio of these velocities is 2.42. i.e., This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air. In other words, the speed of light in diamond is
1/2.42
times the speed of light in vacuum.
Explanation:
a) Draw and label the diagram given :
(i) Incident ray
(ii) Refracted ray
(iii) Emergent ray
(iv) Angle of reflection
(v) Angle of deviation
(v) Angle of emergence
(b) The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?
Answer:
It says about the motion and the graph of the object is stationary, basically travelling at the same speed at any time of the graph. It will never change.
Explanation:
To draw a diagram:
1. Draw an object and represent the speed as stationary and constant at any time.
Answer:
The period of rotation is
T=8.025s
Explanation:
The person is undergoing simple harmonic motion on the wheel
Given data
mass of the person =75kg
Radius of wheel r=16m
Velocity =8.25m/s
The oscillating period of simple harmonic motion is given as
T=(2*pi)/2=2*pi √r/g
Assuming that g=9.81m/s
Substituting our data into the expression we have
T=2*3.142 √ 16/9.81
T=6.284*1.277
T=8.025s
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
I'm going to do the math on the top and then on the bottom and divide at the end.
and now when I divide I will express my answer to the correct number of sig dig's:
6.45 × 10¹⁶ N