Answer:
The displacement is 386.16m
Explanation:
A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement 
R= sqrt(vector1+vector2)
Since this is a right angle triangle 
R= sqrt(248^2 + 296^2)
R= sqrt(149120)
R= 386.16m
Displacement = 386.16m
 
        
             
        
        
        
Answer:
72
Explanation:
The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.
In this problem, the velocity of the sport car is given by the expression

In order to find the expression for the position of the car, we integrate this expression. We find:

where C is an arbitrary constant.
Here we want to find the displacement after 3 seconds. The position at t = 0 is

While the position after t = 3 s is

Therefore, the displacement of the car in 3 seconds is

 
        
             
        
        
        
Answer:
The general equation for conservation of momentum during a collision between n number of objects is given as: [m i ×v i a ] = [m i ×v i b ] Where m i is the mass of object i , v i a is the velocity of object i before the collision, and v i b is the velocity of object i after the collision.
Explanation:
 
        
             
        
        
        
Answer:
 T
 T 
Explanation:
 = Power of the bulb = 100 W
  = Power of the bulb = 100 W
 = distance from the bulb = 2.5 m
  = distance from the bulb = 2.5 m 
 = Intensity of light at the location
 = Intensity of light at the location 
Intensity of the light at the location is given as 


 = 1.28 W/m²
 = 1.28 W/m²
 = maximum magnetic field
 = maximum magnetic field 
Intensity is given as 


 T
 T 
 
        
             
        
        
        
Answer:
 #_photon = 5 10²⁰ photons / s
Explanation:
For this exercise let's calculate the energy of a single quantum of energy, use Planck's law
          E = h f
          c= λ f
          E = h c / λ
           λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m
Let's calculate
           E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹
           E₀ = 19.89 10⁻²⁰ J
This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w
                 #_photon = P / E₀
                #_photon = 100 / 19.89 10⁻²⁰
               #_photon = 5 10²⁰ photons / s