Why would you be surprised if hockey puck started moving without being hit? Why would you be surprised if a fast moving

Monochromatic light from a distant source is incident on a slit 0.705 mm wide. On a screen 2.13 m away, the distance from the ce

vampirchik [111]

Answer:

492.183 nm

Explanation:

x = Distance from the central maximum to the first minimum = 1.35 mm

l = Distance of screen = 2.13 m

d = Distance of gap = 0.705 mm

m = Order = 1

We have the relation

$tan\theta=\dfrac{x}{l}\\\Rightarrow \theta=tan^{-1}\dfrac{1.35\times 10^{-3}}{2.13}\\\Rightarrow \theta=0.04^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{dsin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.705\times 10^{-3}\times sin0.04}{1}\\\Rightarrow \lambda=4.92183\times 10^{-7}=492.183\ nm$

The wavelength of the light is 492.185 nm

5 0

3 years ago

Read 2 more answers

The electron drift speed in a metal wire is exceedingly slow. Yet, when you turnon the light switch the light begins to illumina

elixir [45]

Explanation:

The misunderstanding here is that the thing that turns on the light bulb is not the same electrons near the light switch. So, the electrons near the switch is not moving all the way across the circuit instantly. The electrons are distributed across the wire. When the light switch is turned on, the circuit is connected and there is a potential difference between the bulb and the source. This potential difference creates an electric field, and free electrons move under the influence of this electric field according to Coulomb's Law. When they start to move the electrons closest to the bulb causes the bulb to glow.

So, the important factor here is not the drift velocity of the electrons but the number of electrons and the strength of the electric field.

6 0

3 years ago

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

34kurt

Answer:

The centripetal acceleration of the runner is $1.73\ m/s^2$ .

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

v is the velocity of runner

$v=\dfrac{200\ m}{24\ s}\\\\v=8.34\ m/s$

Centripetal acceleration,

$a=\dfrac{(8.34)^2}{40}\\\\a=1.73\ m/s^2$

So, the centripetal acceleration of the runner is $1.73\ m/s^2$ . Hence, this is the required solution.

5 0

4 years ago

The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00μT. (a) At what distance is it 0.100μ

maw [93]

The distance should be 4m from the wire in order to get the magnetic field of 0.100μ .

The magnitude and direction of the magnetic field due to a straight wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by $B=\frac{u_0I}{2\pi r}$ ...(1)
Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.