The extension of the refracted rays will intersect at a point. This point is known as the focal point. Notice that a diverging lens such as this double concave lens does not really focus the incident light rays that are parallel to the principal axis; rather, it diverges these light rays.

Explanation:

5 0

3 years ago

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A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

blondinia [14]

Answer:

1.2 m/s

0.31 m

0.15 m

Explanation:

Time period is

$T=2.5\times 2\\\Rightarrow T=5\ s$

Frequency is

$f=\dfrac{1}{T}\\\Rightarrow f=\dfrac{1}{5}\\\Rightarrow f=0.2\ Hz$

Velocity is given by

$v=f\lambda\\\Rightarrow v=0.2\times 6\\\Rightarrow v=1.2\ m/s$

The waves are traveling at 1.2 m/s

Amplitude is given by

$A=\dfrac{d}{2}\\\Rightarrow A=\dfrac{0.62}{2}\\\Rightarrow A=0.31\ m$

Amplitude is 0.31 m

If d = 0.3 m

$A=\dfrac{0.3}{2}=0.15\ m$

The amplitude would be 0.15 m. The speed would remain the same.

8 0

4 years ago

A photographer in a helicopter ascending vertically at a constant rate of 11.5 m/s accidentally drops a camera out the window wh

Degger [83]

The problem is a free fall problem as the camera drops. The vertical velocity is zero. It is the action of gravity that takes place here. To calculate the time it takes to reach the ground, we do as follows:

t = √2y/g = √(2)(70)/9.81 = 3.78 s

The velocity as it impacts the ground would be,

v = √2gy = 37.04 m/s

Hope this helps.

6 0

4 years ago

Consider the parallel-plate capacitor shown in the figure. The plate separation is 1.8 mm and the the electric field inside is 2

quester [9]

We have that the Velocities is mathematically given as

a)v=2.8e5m/s

b)v_2=3.11e5m/s

From the question we are told

Consider the parallel-plate <u>capacitor </u>shown in the figure.
The plate separation is 1.8 mm and the the electric field inside is 23 N/C.
An electron is positioned halfway between the plates and is given some initial velocity, vi.

<h3>Velocity</h3>

Generally the equation for the workdone is mathematically given as

$-qEr=0.5m(u/2)^2\\\\Therefore\\\\1.6e-19*19*2.9e-3=0.5*9.1e-31*v^2/4$

v=2.8e5m/s

applying work

$-qEr=0.5m(u/2)^2\\\\1.6e-19*19*2.9e-3=0.5*9.1e-31(v_2^2-7.75e10)$

v_2=3.11e5m/s

For more information on Velocity visit

brainly.com/question/7359669

8 0

3 years ago