Answer: Option D: 5.5×10²Joules
Explanation:
Work done is the product of applied force and displacement of the object in the direction of force.
W = F.s = F s cosθ
It is given that the force applied is, F = 55 N
The displacement in the direction of force, s = 10 m
The angle between force and displacement, θ = 0°
Thus, work done on the object:
W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J
Hence, the correct option is D.
The law of conservation of energy<span>, a fundamental concept of physics, states that the total amount of </span>energy<span> remains constant in an isolated system. It implies that </span>energy<span> can neither be created nor destroyed, but can be change from one form to another.</span>
Phenyl oxalate ester is responsible for the luminescence in aglow stick<span>. The reaction with hydrogen peroxide </span>causes<span> the liquid inside a </span><span>glow stick to glow</span>
To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then


Here,
= Mass of each object
= Initial velocity of each object
= Final velocity of each object
When they position the final velocities of the bodies it is the same and the car is stationary then,

Rearranging to find the final velocity



The expression for the impulse received by the first car is


Replacing,


The negative sign show the opposite direction.
Answer:
a) Diffusion coefficient, D = 1.5 in/hr
b) Mean jump frequency, f = 0.0833 Hz
Explanation:
a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:
..........(1)
Where <r> = mean displacement
D = Diffusion coefficient
t = time = 12 hrs
sum of the squares of the distance divided by 100 is 36 in2.
<r>²= 36 in²
Substituting these values into equation (1) above

b) Mean jumping distance, <r> = 0.1 inches
Applying equation (1) again
Where D = 1.5 in/hr


The mean jump frequency, f = 1/t
f = 1/12
f = 0.0833 Hz