Answer:
Explanation:
As we know that the resultant of two vectors is given as
here we know that
now we have
Answer:
Explanation:
speed of bat = 6 m/s
sound wave frequency emitted by bat = 30.0 kHz
as we know,
speed of sound (c)= 343 m/s
now frequency received by bat is equal to
hence the frequency hear by bat will be 29.98 Hz
Answer:
486nm
Explanation:
in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that
![\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bwavelength%7D%3DR.%5B%5Cfrac%7B1%7D%7Bn_%7Bf%7D%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%7D%7Bn_%7Bi%7D%5E%7B2%7D%20%7D%5D%20%5C%5Cn_%7Bf%7D%3D2%5C%5Cn_%7Bi%7D%3D4%5C%5CR%3DRydberg%20constant%20%3D1.097%2A10%5E%7B7%7Dm%5E%7B-1%7D%5C%5Csubtitiute%20%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D1.097%2A10%5E%7B7%7D%5B%5Cfrac%7B1%7D%7B2%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%7D%7B4%5E%7B2%7D%7D%5D%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D%201.097%2A10%5E%7B7%7D%2A0.1875%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D%202.06%2A10%5E%7B6%7D%5C%5Cwavelength%3D4.86%2A10%7B-7%7Dm%5C%5Cwavelength%3D%20486nm%5C%5C)
Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state
Explanation:
q= n e
6 × 10^-11 = n (1.6 × 10^-19)
n = 6×10^-11 / 1.6 × 10^-19
n= 3.75 × 10⁸ electrons
Answer:
59.4 meters
Explanation:
The correct question statement is :
A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?
Solution:
We know for a circle of radius r and θ angle by an arc of length S at the center,
S=rθ
This gives
θ=S/r
also we know angular velocity
ω=θ/t where t is time
or
θ=ωt
and we know
1 revolution =2π radians
From this we have
angular velocity ω = 1.4 revolutions per sec = 1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec
Putting values of ω and time t in
θ=ωt
we have
θ= 8.8 rad / sec × 4.5 sec
θ= 396 radians
We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)
put this value of θ and r in
S=rθ
we have
S= 396 radians ×0.15 m=59.4 m
