The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW
Answer: the airy pattern can only arise from wave propagation
Explanation:if particles went in straight lines through a slit, they would progate linearly and not interfere. The airy pattern arises from diffraction as waves interfere, producing peaks (constructive interference where peaks of waves from each slit coincide) and troughs (destructive interference where peaks and troughs of waves from each slit cancel out). If intensity rather than field is measured nodes occur where 0 values line up instead of troughs
Answer: I never said anything to them but if i did it would be "I need a break from you, i love you but sometimes I have to find the path on my own." i would say this if i wasn't scared to hurt them.
Explanation:
<span>Answer:
sin(incidence)/sin(refraction) = n_refraction/n_incidence
sin(50) / sin(x) = 1.5 / 1
sin(50)/1.5 = sin(x)
sin(x) = 0.511
x = 30.71o
B]
50 degrees, same as the angle going in.
You can show that by reversing the steps in A.
sin(30.7)/sin(x) = 1/1.5
C]
The glass is 5 cm thick.
The reference angle = 30.7o
Tan(30.7) = displacement / thickness
Tan(30.7) = x / 5
5*sin(30.7) = x
x = 2.97 cm which is the displacement.</span>
The total flux through the cylinder is zero.
In fact, the electric flux through a surface (for a uniform electric field) is given by:
where
E is the intensity of the electric field
A is the surface
is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.
We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because 
and 
).
On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be
where r is the radius, and where we have taken 
since the perpendicular to the surface is parallel to the direction of the electric field, so 
. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so 
and 
, therefore the flux is
And the net flux through the cylinder is
