V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s
v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm
Answer:
4
Explanation:
In order for the current to continue flowing through the circuit (and for the bulbs to continue shining), there must be a closed path containing the battery where current can flow. Let's see the effect of removing each bulb on the circuit:
- 1: when removing bulb 1 only, the current can still flow through the path battery-bulb 3- bulb 4
- 2: when removing bulb 2 only, the current can still flow through the path battery-bulb 3- bulb 4
- 3: when removing bulb 3 only, the current can still flow through the path battery-bulb 1-bulb 2- bulb 4
- 4: when removing bulb 4 only, the current can no longer flow. In fact, there is no closed path that contains the battery now, so the current will not flow and all the bulbs will stop shining.
Answer:
Frequency = 3.19 * 10^14 Hz or 1/s
Explanation:
Relationship b/w frequency and wavelength can be expressed as:
C = wavelength * frequency, where c is speed of light in vacuum which is 3.0*10^8 m/s.
Now simply input value (but before that convert wavelength into meters to match the units, you do this by multiply it by 10^-9 so it will be 940*10^-9)
3.0 * 10^8 = Frequency * 940 x 10^-9
Frequency = 3.19 * 10^14 Hz or 1/s
Answer:
(A). The order of the bright fringe is 6.
(B). The width of the bright fringe is 3.33 μm.
Explanation:
Given that,
Fringe width d = 0.5 mm
Wavelength = 589 nm
Distance of screen and slit D = 1.5 m
Distance of bright fringe y = 1 cm
(A) We need to calculate the order of the bright fringe
Using formula of wavelength
![\lambda=\dfrac{dy}{mD}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bdy%7D%7BmD%7D)
![m=\dfrac{d y}{\lambda D}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7Bd%20y%7D%7B%5Clambda%20D%7D)
Put the value into the formula
![m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B1%5Ctimes10%5E%7B-2%7D%5Ctimes0.5%5Ctimes10%5E%7B-3%7D%7D%7B589%5Ctimes10%5E%7B-9%7D%5Ctimes1.5%7D)
![m=5.65 = 6](https://tex.z-dn.net/?f=m%3D5.65%20%3D%206)
(B). We need to calculate the width of the bright fringe
Using formula of width of fringe
![\beta=\dfrac{yd}{D}](https://tex.z-dn.net/?f=%5Cbeta%3D%5Cdfrac%7Byd%7D%7BD%7D)
Put the value in to the formula
![\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}](https://tex.z-dn.net/?f=%5Cbeta%3D%5Cdfrac%7B1%5Ctimes10%5E%7B-2%7D%5Ctimes0.5%5Ctimes10%5E%7B-3%7D%7D%7B1.5%7D)
![\beta=3.33\times10^{-6}\ m](https://tex.z-dn.net/?f=%5Cbeta%3D3.33%5Ctimes10%5E%7B-6%7D%5C%20m)
![\beta=3.33\ \mu m](https://tex.z-dn.net/?f=%5Cbeta%3D3.33%5C%20%5Cmu%20m)
Hence, (A). The order of the bright fringe is 6.
(B). The width of the bright fringe is 3.33 μm.
As per Newton's III law every action has equal and opposite reaction
So here we can say that
every body which apply force on other body must have a reaction force of same magnitude in opposite direction
So here if ball hits the ground by 50 N force then the ball must have a reaction force on itself with same magnitude and opposite direction
the magnitude of the force will be 50 N
and its direction is opposite to the force that ball apply on the floor