Answer:
The value is 
Explanation:
Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

Here u = 0 \ m/s
a = 13.5 
So
=>
The is distance it covers at this time is

=> 
=> 
Now when sled stops its the final velocity is
while the initial velocity will be the velocity after its acceleration i.e
So

Here
, the negative sign shows that it is deceleration
So

=> 
Answer:
a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m
Explanation:
The diffraction network is described by the expression
d sin θ= m λ
Where m corresponds to the diffraction order
Let's use trigonometry to find the breast
tan θ = y / L
The diffraction spectrum is measured at very small angles, therefore
tan θ = sin θ / cos θ = sin θ
We replace
d y / L = m λ
Let's place in the first order m = 1
Let's look for the separation of the lines (d)
d = λ L / y
d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²
d = 332.33 10⁻⁹ m
Now we can look for the wavelength of the other line
λ = d y / L
λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²
λ = 189.43 10⁻⁹ m
Part B
The compound wavelength B
λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²
λ = 269.19 10⁻⁹ m
A & B
Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.
Explanation:
Real images are formed where the rays converge, a property of images by convex lenses. Convex lenses can be used to magnify objects. If the image occurs before the focal point of the lens then the image will be upright but smaller. The images inverts and gets bigger past the focal point.
Virtual images are property of concave lenses. These images appear closer but smaller than the real object.
Learn More:
For more on images formed by lenses check out;
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Answer:
66.98 db
Explanation:
We know that

L_T= Total signal level in db
n= number of sources
L_S= signal level from signal source.

= 66.98 db
Correct option: A
An object remains at rest until a force acts on it.
As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.