Two colours of light which mix to make white light are called

A series of pulses, each of amplitude 0.1m , is sent down a string that is attached to a post at one end. The pulses are reflect

sp2606 [1]

The net displacement at a point on the string where the pulses cross is 0.2 m.

The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.

A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.

At the post, the pulses are reflected and return along the string without losing any of their amplitude.

Now, let's say the ends are free.

There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.

Now, since A = 0.1 m

Then, 2A = 2(0.1) = 0.2 m

As a result, the net displacement at the string's intersection of two pulses is 0.2 m.

The correct option is (c).

Learn more about amplitude here:

brainly.com/question/3613222

#SPJ4

4 0

2 years ago

Bambi the young dear was distracted Buy butterfly and jumped into the road in front of the two vehicles as shown in the diagram

bagirrra123 [75]

Speed of car A is given as

$v_a = 70 mph$

now we need to convert it into SI units

1 miles = 1609 m

1 hour = 3600 s

now we have

$v_a = 70 *\frac{1609}{3600} = 31.3 m/s$

now its distance from Bambi is given as

$d_a = 350 m$

time taken by it to hit the Bambi

$t = \frac{d}{v}$

$t = \frac{350}{31.3}$

$t = 11.2 s$

Now other car is moving at speed 50 mph

so its speed in SI unit will be

$v_b = 50* \frac{1609}{3600}$

$v_b = 22.35 m/s$

now its distance from Bambi is given as

$d_b = 590 feet$

as we know that 1 feet = 0.3048 m

$d_b = 590*0.3048 = 179.83 m$

now the time to hit the other car is

$t_2 = \frac{179.83}{22.35}$

$t_b = 8.05 s$

So Car B will hit the Bambi first

7 0

3 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$