Answer:
Approximately 0.0898 W/m².
Explanation:
The intensity of light measures the power that the light delivers per unit area.
The source in this question delivers a constant power of 
. If the source here is a point source, that of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.
The surface area of a sphere of radius 
is equal to 
. For the imaginary 9.6-meter sphere here, the surface area will be:
.
That power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:
.
To the Earth in less than ten minutes.
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
M=2.45 because you multiply out the equation on the right and divide by 10
Answer:
The average velocity is
and 
respectively.
Explanation:
Let's start writing the vertical position equation :
Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt = 
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :
For the position variation we use the vertical position equation :
Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is
For the second time interval :
t1 = 4 s → t2 = 9 s
Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :
Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then
The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is
